APPLICATIONS OF THE DEFINITE INTEGRAL IN GEOMETRY, SCIENCE, AND ENGINEERING

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1 6 Courtes NASA APPLICATIONS OF THE DEFINITE INTEGRAL IN GEOMETRY, SCIENCE, AND ENGINEERING Clculus is essentil for the computtions required to lnd n stronut on the Moon. In the lst chpter we introduced the definite integrl s the limit of Riemnn sums in the contet of finding res. However, Riemnn sums nd definite integrls hve pplictions tht etend fr beond the re problem. In this chpter we will show how Riemnn sums nd definite integrls rise in such problems s finding the volume nd surfce re of solid, finding the length of plne curve, clculting the work done b force, finding the center of grvit of plnr region, finding the pressure nd force eerted b fluid on submerged object, nd finding properties of suspended cbles. Although these problems re diverse, the required clcultions cn ll be pproched b the sme procedure tht we used to find res breking the required clcultion into smll prts, mking n pproimtion for ech prt, dding the pproimtions from the prts to produce Riemnn sum tht pproimtes the entire quntit to be clculted, nd then tking the limit of the Riemnn sums to produce n ect result. 6. AREA BETWEEN TWO CURVES In the lst chpter we showed how to find the re between curve = f() nd n intervl on the -is. Here we will show how to find the re between two curves. Δ k = f() A REVIEW OF RIEMANN SUMS Before we consider the problem of finding the re between two curves it will be helpful to review the bsic principle tht underlies the clcultion of re s definite integrl. Recll tht if f is continuous nd nonnegtive on [,b], then the definite integrl for the re A under = f() over the intervl [,b] is obtined in four steps (Figure 6..): f (* k ) Figure 6.. * k b Divide the intervl [,b] into n subintervls, nd use those subintervls to divide the region under the curve = f() into n strips. Assuming tht the width of the kth strip is k, pproimte the re of tht strip b the re f(k ) k of rectngle of width k nd height f(k ), where k is point in the kth subintervl. Add the pproimte res of the strips to pproimte the entire re A b the Riemnn sum: n A f(k ) k k= 43

2 44 Chpter 6 / Applictions of the Definite Integrl in Geometr, Science, nd Engineering n k = b f (* k ) f() Δ k d Effect of the limit process on the Riemnn sum Figure 6.. Tke the limit of the Riemnn sums s the number of subintervls increses nd ll their widths pproch zero. This cuses the error in the pproimtions to pproch zero nd produces the following definite integrl for the ect re A: A = lim m k n f(k ) k = k= b f()d Figure 6.. illustrtes the effect tht the limit process hs on the vrious prts of the Riemnn sum: The quntit k in the Riemnn sum becomes the vrible in the definite integrl. The intervl width k in the Riemnn sum becomes the d in the definite integrl. The intervl [,b], which is the union of the subintervls with widths,,..., n, does not pper eplicitl in the Riemnn sum but is represented b the upper nd lower limits of integrtion in the definite integrl. AREA BETWEEN = f () AND = g() We will now consider the following etension of the re problem. 6.. first re problem Suppose tht f nd g re continuous functions on n intervl [,b] nd f() g() for b [This mens tht the curve = f()lies bove the curve = g() nd tht the two cn touch but not cross.] Find the re A of the region bounded bove b = f(), below b = g(), nd on the sides b the lines = nd = b (Figure 6..3). = f() Δ k = f() A b f (* k ) g(* k ) b Figure 6..3 = g() () = g() (b) * k To solve this problem we divide the intervl [,b] into n subintervls, which hs the effect of subdividing the region into n strips (Figure 6..3b). If we ssume tht the width of the kth strip is k, then the re of the strip cn be pproimted b the re of rectngle of width k nd height f(k ) g( k ), where k is point in the kth subintervl. Adding these pproimtions ields the following Riemnn sum tht pproimtes the re A: n A [f(k ) g( k )] k k= Tking the limit s n increses nd the widths of ll the subintervls pproch zero ields the following definite integrl for the re A between the curves: n b A = lim [f(k ) m k g( k )] k = [f() g()] d k=

3 6. Are Between Two Curves 45 In summr, we hve the following result = re formul If f nd g re continuous functions on the intervl [,b], nd if f() g() for ll in [,b], then the re of the region bounded bove b = f(), below b = g(), on the left b the line =, nd on the right b the line = b is A = b [f() g()] d () Figure 6..4 = Emple Find the re of the region bounded bove b = + 6, bounded below b =, nd bounded on the sides b the lines = nd =. Solution. The region nd cross section re shown in Figure The cross section etends from g() = on the bottom to f() = + 6 on the top. If the cross section is moved through the region, then its leftmost position will be = nd its rightmost position will be =. Thus, from () [ A = [( + 6) ] 3 ] d = + 6 = = 34 3 Wht does the integrl in () represent if the grphs of f nd g cross ech other over the intervl [,b]? How would ou find the re between the curves in this cse? It is possible tht the upper nd lower boundries of region m intersect t one or both endpoints, in which cse the sides of the region will be points, rther thn verticl line segments (Figure 6..5). When tht occurs ou will hve to determine the points of intersection to obtin the limits of integrtion. = f() = f() = g() b = g() b Figure 6..5 The left-hnd boundr reduces to point. Both side boundries reduce to points. = (3, 9) (, 4) 4 = Figure 6..6 Emple Find the re of the region tht is enclosed between the curves = nd = + 6. Solution. A sketch of the region (Figure 6..6) shows tht the lower boundr is = nd the upper boundr is = + 6. At the endpoints of the region, the upper nd lower boundries hve the sme -coordintes; thus, to find the endpoints we equte This ields from which we obtin = nd = + 6 () = + 6 or 6 = or ( + )( 3) = = nd = 3 Although the -coordintes of the endpoints re not essentil to our solution, the m be obtined from () b substituting = nd = 3 in either eqution. This ields = 4 nd = 9, so the upper nd lower boundries intersect t (, 4) nd (3, 9).

4 46 Chpter 6 / Applictions of the Definite Integrl in Geometr, Science, nd Engineering From () with f() = + 6, g() =, =, nd b = 3, we obtin the re 3 [ A = [( + 6) ] 3 3 ] d = + 6 = 7 ( 3 ) = In the cse where f nd g re nonnegtive on the intervl [,b], the formul A = b [f() g()] d = b f()d b g() d sttes tht the re A between the curves cn be obtined b subtrcting the re under = g() from the re under = f() (Figure 6..7). = f () = f() = f() A = = g() b = g() b = g() b Are between f nd g Are below f Are below g Figure 6..7 v v = v (t) A Cr Cr Emple 3 Figure 6..8 shows velocit versus time curves for two rce crs tht move long stright trck, strting from rest t the sme time. Give phsicl interprettion of the re A between the curves over the intervl t T. Figure 6..8 v = v (t) T t Solution. From () T T T A = [v (t) v (t)] dt = v (t) dt v (t) dt Since v nd v re nonnegtive functions on [,T], it follows from Formul (4) of Section 5.7 tht the integrl of v over [,T] is the distnce trveled b cr during the time intervl t T, nd the integrl of v over [,T] is the distnce trveled b cr during the sme time intervl. Since v (t) v (t) on [,T], cr trvels frther thn cr does over the time intervl t T, nd the re A represents the distnce b which cr is hed of cr t time T. Some regions m require creful thought to determine the integrnd nd limits of integrtion in (). Here is sstemtic procedure tht ou cn follow to set up this formul. It is not necessr to mke n etremel ccurte sketch in Step ; the onl purpose of the sketch is to determine which curve is the upper boundr nd which is the lower boundr. Finding the Limits of Integrtion for the Are Between Two Curves Step. Sketch the region nd then drw verticl line segment through the region t n rbitrr point on the -is, connecting the top nd bottom boundries (Figure 6..9). Step. The -coordinte of the top endpoint of the line segment sketched in Step will be f(), the bottom one g(), nd the length of the line segment will be f() g(). This is the integrnd in (). Step 3. To determine the limits of integrtion, imgine moving the line segment left nd then right. The leftmost position t which the line segment intersects the region is = nd the rightmost is = b (Figures 6..9b nd 6..9c).

5 6. Are Between Two Curves 47 f() f() g() g() b b Figure 6..9 () (b) (c) There is useful w of thinking bout this procedure: If ou view the verticl line segment s the cross section of the region t the point, then Formul () sttes tht the re between the curves is obtined b integrting the length of the cross section over the intervl [,b]. It is possible for the upper or lower boundr of region to consist of two or more different curves, in which cse it will be convenient to subdivide the region into smller pieces in order to ppl Formul (). This is illustrted in the net emple. Emple 4 Find the re of the region enclosed b = nd =. (4, ) = = A ( = + ) 4 (, ) () (4, ) = = A A 4 (, ) = (b) Figure 6.. Solution. To determine the pproprite boundries of the region, we need to know where the curves = nd = intersect. In Emple we found intersections b equting the epressions for. Here it is esier to rewrite the ltter eqution s = + nd equte the epressions for, nmel, = nd = + (3) This ields = + or = or ( + )( ) = from which we obtin =, =. Substituting these vlues in either eqution in (3) we see tht the corresponding -vlues re = nd = 4, respectivel, so the points of intersection re (, ) nd (4, ) (Figure 6..). To ppl Formul (), the equtions of the boundries must be written so tht is epressed eplicitl s function of. The upper boundr cn be written s = (rewrite = s =± nd choose the + for the upper portion of the curve). The lower boundr consists of two prts: = for nd = for 4 (Figure 6..b). Becuse of this chnge in the formul for the lower boundr, it is necessr to divide the region into two prts nd find the re of ech prt seprtel. From () with f() =,g() =, =, nd b =, we obtin A = [ ( [ ] )] d = d= 3 3/ = 4 3 = 4 3 From () with f() =,g() =, =, nd b = 4, we obtin A = = 4 [ ( )] d = [ 3 3/ + ] 4 = 4 ( + )d ( ) ( 3 ) + = 9 6

6 48 Chpter 6 / Applictions of the Definite Integrl in Geometr, Science, nd Engineering Thus, the re of the entire region is A = A + A = = 9 REVERSING THE ROLES OF AND Sometimes it is much esier to find the re of region b integrting with respect to rther thn. We will now show how this cn be done. d = w() 6..3 second re problem Suppose tht w nd v re continuous functions of on n intervl [c, d] nd tht = v() c Figure 6.. w() v() for c d [This mens tht the curve = w() lies to the right of the curve = v() nd tht the two cn touch but not cross.] Find the re A of the region bounded on the left b = v(), on the right b = w(), nd bove nd below b the lines = d nd = c (Figure 6..). Proceeding s in the derivtion of (), but with the roles of nd reversed, leds to the following nlog of re formul If w nd v re continuous functions nd if w() v() for ll in [c, d], then the re of the region bounded on the left b = v(), on the right b = w(), below b = c, nd bove b = d is d A = d c [w() v()] d (4) v() c Figure 6.. w() The guiding principle in ppling this formul is the sme s with (): The integrnd in (4) cn be viewed s the length of the horizontl cross section t n rbitrr point on the -is, in which cse Formul (4) sttes tht the re cn be obtined b integrting the length of the horizontl cross section over the intervl [c, d] on the -is (Figure 6..). In Emple 4, we split the region into two prts to fcilitte integrting with respect to. In the net emple we will see tht splitting this region cn be voided if we integrte with respect to. Emple 5 Find the re of the region enclosed b = nd =, integrting with respect to. The choice between Formuls () nd (4) is usull dictted b the shpe of the region nd which formul requires the lest mount of splitting. However, sometimes one might choose the formul tht requires more splitting becuse it is esier to evlute the resulting integrls. Solution. As indicted in Figure 6.. the left boundr is =, the right boundr is =, nd the region etends over the intervl. However, to ppl (4) the equtions for the boundries must be written so tht is epressed eplicitl s function of. Thus, we rewrite = s = +. It now follows from (4) tht [ A = [( + ) ] 3 ] d = + = 9 3 which grees with the result obtined in Emple 4.

7 6. Are Between Two Curves 49 QUICK CHECK EXERCISES 6. (See pge 4 for nswers.). An integrl epression for the re of the region between the curves = 3 nd = e nd bounded on the sides b = nd = is.. An integrl epression for the re of the prllelogrm bounded b = + 8, = 3, =, nd = 5 is. The vlue of this integrl is. 3. () The points of intersection for the circle + = 4 nd the line = + re nd. (b) Epressed s definite integrl with respect to, gives the re of the region inside the circle + = 4 nd bove the line = +. (c) Epressed s definite integrl with respect to, gives the re of the region described in prt (b). 4. The re of the region enclosed b the curves = nd = 3 is. EXERCISE SET 6. Grphing Utilit C CAS 4 Find the re of the shded region.. = = = 4. = = / = 4 = 4 = 5 6 Find the re of the shded region b () integrting with respect to nd (b) integrting with respect to = = (, 4) 5 = 4 (4, 4) (, ) = Sketch the region enclosed b the curves nd find its re. 7. =,=, = 4,= 8. = 3 4, =, =, = 9. = cos, =, = π/4, = π/. = sec, =, = π/4, = π/4. = sin, =, = π/4, = 3π/4. =, = 5 3. = e,= e,=, = ln 4. = /, =, =, = e 5. = +,= 6. =, = 7. = +, = =, = 4, = Use grphing utilit, where helpful, to find the re of the region enclosed b the curves. 9. = , =. = 3,= 3. = sin, = cos, =, = π. = 3 4, = 3. = 3, = 4. = , = 5. = e,= 6. = (ln ),= True Flse Determine whether the sttement is true or flse. Eplin our nswer. [In ech eercise, ssume tht f nd g re distinct continuous functions on [,b] nd tht A denotes the re of the region bounded b the grphs of = f(), = g(), =, nd = b.] 7. If f nd g differ b positive constnt c, then A = c(b ). 8. If b [f() g()] d = 3 then A = If b [f() g()] d = then the grphs of = f() nd = g() cross t lest once on [,b]. 3. If b A = [f() g()] d then the grphs of = f() nd = g() don t cross on [,b].

8 4 Chpter 6 / Applictions of the Definite Integrl in Geometr, Science, nd Engineering C C C 3. Estimte the vlue of k( <k<) so tht the region enclosed b = /, =, =, nd = k hs n re of squre unit. 3. Estimte the re of the region in the first qudrnt enclosed b = sin nd = sin. 33. Use CAS to find the re enclosed b = 3 nd = Use CAS to find the ect re enclosed b the curves = nd = Find horizontl line = k tht divides the re between = nd = 9 into two equl prts. 36. Find verticl line = k tht divides the re enclosed b =, =, nd = into two equl prts. 37. () Find the re of the region enclosed b the prbol = nd the -is. (b) Find the vlue of m so tht the line = m divides the region in prt () into two regions of equl re. 38. Find the re between the curve = sin nd the line segment joining the points (, ) nd (5π/6, /) on the curve Use Newton s Method (Section 4.7), where needed, to pproimte the -coordintes of the intersections of the curves to t lest four deciml plces, nd then use those pproimtions to pproimte the re of the region. 39. The region tht lies below the curve = sin nd bove the line =., where. 4. The region enclosed b the grphs of = nd = cos. 4. The region enclosed b the grphs of = (ln )/ nd =. 4. The region enclosed b the grphs of = 3 cos nd = /( + ). 43. Find the re of the region tht is enclosed b the curves = nd = sin. 44. Referring to the ccompning figure, use CAS to estimte the vlue of k so tht the res of the shded regions re equl. Source: This eercise is bsed on ProblemA tht ws posed in the Fift-Fourth Annul Willim Lowell Putnm Mthemticl Competition. = sin = k FOCUS ON CONCEPTS c Figure E Two rcers in djcent lnes move with velocit functions v (t) m/s nd v (t) m/s, respectivel. Suppose tht the rcers re even t time t = 6 s. Interpret the vlue of the integrl in this contet. 6 [v (t) v (t)] dt 46. The ccompning figure shows ccelertion versus time curves for two crs tht move long stright trck, ccelerting from rest t the strting line. Wht does the re A between the curves over the intervl t T represent? Justif our nswer. = (t) Cr Cr = (t) t T Figure E The curves in the ccompning figure model the birth rtes nd deth rtes (in millions of people per er) for countr over 5-er period. Wht does the re A between the curves over the intervl [96, ] represent? Justif our nswer. Millions/er Birth rte Deth rte Yer Figure E The ccompning figure shows the rte t which trnsderml mediction is bsorbed into the bloodstrem of n individul, s well s the rte t which the mediction is eliminted from the bloodstrem b metboliztion. Both rtes re in units of microgrms per hour (μg/h) nd re displed over n 8-hour period. Wht does the re A between the curves over the intervl [, 8] represent? Justif our nswer. mg/h Absorption rte Elimintion rte Hours Figure E Find the re of the region enclosed between the curve / + / = / nd the coordinte es. 5. Show tht the re of the ellipse in the ccompning figure is πb. [Hint: Use formul from geometr.] + b = b Figure E-5

9 6. Volumes b Slicing; Disks nd Wshers 4 5. Writing Suppose tht f nd g re continuous on [,b] but tht the grphs of = f() nd = g() cross severl times. Describe step-b-step procedure for determining the re bounded b the grphs of = f(), = g(), =, nd = b. QUICK CHECK ANSWERS 6.. (c) [( 3 ) e ] d. 5 [( ) + 4 ] d 4. [( + 8) ( 3)] d; Writing Suppose tht R nd S re two regions in the plne tht lie between pir of lines L nd L tht re prllel to the -is. Assume tht ech line between L nd L tht is prllel to the -is intersects R nd S in line segments of equl length. Give n informl rgument tht the re of R is equl to the re of S. (Mke resonble ssumptions bout the boundries of R nd S.) 3. () (, ); (, ) (b) [ 4 ( + )] d 6. VOLUMES BY SLICING; DISKS AND WASHERS In the lst section we showed tht the re of plne region bounded b two curves cn be obtined b integrting the length of generl cross section over n pproprite intervl. In this section we will see tht the sme bsic principle cn be used to find volumes of certin three-dimensionl solids. VOLUMES BY SLICING Recll tht the underling principle for finding the re of plne region is to divide the region into thin strips, pproimte the re of ech strip b the re of rectngle, dd the pproimtions to form Riemnn sum, nd tke the limit of the Riemnn sums to produce n integrl for the re. Under pproprite conditions, the sme strteg cn be used to find the volume of solid. The ide is to divide the solid into thin slbs, pproimte the volume of ech slb, dd the pproimtions to form Riemnn sum, nd tke the limit of the Riemnn sums to produce n integrl for the volume (Figure 6..). Sphere cut into horizontl slbs Right prmid cut into horizontl slbs Right circulr cone cut into horizontl slbs Right circulr cone cut into verticl slbs Figure 6.. Cross section In thin slb, the cross sections do not vr much in size nd shpe. Figure 6.. Wht mkes this method work is the fct tht thin slb hs cross section tht does not vr much in size or shpe, which, s we will see, mkes its volume es to pproimte (Figure 6..). Moreover, the thinner the slb, the less vrition in its cross sections nd the better the pproimtion. Thus, once we pproimte the volumes of the slbs, we cn set up Riemnn sum whose limit is the volume of the entire solid. We will give the detils shortl, but first we need to discuss how to find the volume of solid whose cross sections do not vr in size nd shpe (i.e., re congruent). One of the simplest emples of solid with congruent cross sections is right circulr clinder of rdius r, since ll cross sections tken perpendiculr to the centrl is re circulr regions of rdius r. The volume V of right circulr clinder of rdius r nd height h cn be epressed in terms of the height nd the re of cross section s V = πr h =[re of cross section] [height] ()

10 4 Chpter 6 / Applictions of the Definite Integrl in Geometr, Science, nd Engineering This is specil cse of more generl volume formul tht pplies to solids clled right clinders. A right clinder is solid tht is generted when plne region is trnslted long line or is tht is perpendiculr to the region (Figure 6..3). Some Right Clinders Trnslted squre Trnslted disk Trnslted nnulus Trnslted tringle Figure 6..3 Are A h Volume = A. h Figure 6..4 If right clinder is generted b trnslting region of re A through distnce h, then h is clled the height (or sometimes the width) of the clinder, nd the volume V of the clinder is defined to be V = A h =[re of cross section] [height] () (Figure 6..4). Note tht this is consistent with Formul () for the volume of right circulr clinder. We now hve ll of the tools required to solve the following problem. Cross section S 6.. problem Let S be solid tht etends long the -is nd is bounded on the left nd right, respectivel, b the plnes tht re perpendiculr to the -is t = nd = b (Figure 6..5). Find the volume V of the solid, ssuming tht its cross-sectionl re A() is known t ech in the intervl [,b]. b Cross section re = A() Figure 6..5 To solve this problem we begin b dividing the intervl [,b] into n subintervls, thereb dividing the solid into n slbs s shown in the left prt of Figure If we ssume tht the width of the kth subintervl is k, then the volume of the kth slb cn be pproimted b the volume A(k ) k of right clinder of width (height) k nd cross-sectionl re A(k ), where k is point in the kth subintervl (see the right prt of Figure 6..6). Figure 6..6 S S S S n... n b The cross section here hs re A(* k ). * k Δ k S k Adding these pproimtions ields the following Riemnn sum tht pproimtes the volume V : n V A(k ) k k=

11 6. Volumes b Slicing; Disks nd Wshers 43 Tking the limit s n increses nd the widths of ll the subintervls pproch zero ields the definite integrl n b V = lim A(k ) k = A() d m k k= In summr, we hve the following result. It is understood in our clcultions of volume tht the units of volume re the cubed units of length [e.g., cubic inches (in 3 ) or cubic meters (m 3 )]. 6.. volume formul Let S be solid bounded b two prllel plnes perpendiculr to the -is t = nd = b. If, for ech in [,b], the cross-sectionl re of S perpendiculr to the -is is A(), then the volume of the solid is V = b A() d (3) provided A() is integrble. There is similr result for cross sections perpendiculr to the -is volume formul Let S be solid bounded b two prllel plnes perpendiculr to the -is t = c nd = d. If, for ech in [c, d], the cross-sectionl re of S perpendiculr to the -is is A(), then the volume of the solid is provided A() is integrble. V = d c A() d (4) B(, h) -is In words, these formuls stte: The volume of solid cn be obtined b integrting the cross-sectionl re from one end of the solid to the other. B h O s () (b) Figure 6..7 O C, C h -is Emple Derive the formul for the volume of right prmid whose ltitude is h nd whose bse is squre with sides of length. Solution. As illustrted in Figure 6..7, we introduce rectngulr coordinte sstem in which the -is psses through the pe nd is perpendiculr to the bse, nd the -is psses through the bse nd is prllel to side of the bse. At n in the intervl [,h] on the -is, the cross section perpendiculr to the - is is squre. If s denotes the length of side of this squre, then b similr tringles (Figure 6..7b) s = h h Thus, the re A() of the cross section t is or s = (h ) h A() = s = (h ) h

12 44 Chpter 6 / Applictions of the Definite Integrl in Geometr, Science, nd Engineering nd b (4) the volume is V = h A() d = h = h h (h ) d = h h [ ] h 3 (h )3 = = h Tht is, the volume is of the re of the bse times the ltitude. 3 (h ) d [ + 3 ] h3 = 3 h SOLIDS OF REVOLUTION A solid of revolution is solid tht is generted b revolving plne region bout line tht lies in the sme plne s the region; the line is clled the is of revolution. Mn fmilir solids re of this tpe (Figure 6..8). Some Fmilir Solids of Revolution Ais of revolution Figure 6..8 Right circulr clinder Solid sphere Solid cone Hollowed right circulr clinder () (b) (c) (d) VOLUMES BY DISKS PERPENDICULAR TO THE -AXIS We will be interested in the following generl problem problem Let f be continuous nd nonnegtive on [,b], nd let R be the region tht is bounded bove b = f(), below b the -is, nd on the sides b the lines = nd = b (Figure 6..9). Find the volume of the solid of revolution tht is generted b revolving the region R bout the -is. = f() R b f() b Figure 6..9 () (b)

13 6. Volumes b Slicing; Disks nd Wshers 45 We cn solve this problem b slicing. For this purpose, observe tht the cross section of the solid tken perpendiculr to the -is t the point is circulr disk of rdius f() (Figure 6..9b). The re of this region is A() = π[f()] Thus, from (3) the volume of the solid is = 4 V = b π[f()] d (5) Becuse the cross sections re disk shped, the ppliction of this formul is clled the method of disks. Emple Find the volume of the solid tht is obtined when the region under the curve = over the intervl [, 4] is revolved bout the -is (Figure 6..). Figure 6.. Solution. From (5), the volume is + = r V = b π[f()] d = 4 ] 4 π d = π = 8π π = 5π r r Figure 6.. = f() R = g() b () Emple 3 Derive the formul for the volume of sphere of rdius r. Solution. As indicted in Figure 6.., sphere of rdius r cn be generted b revolving the upper semicirculr disk enclosed between the -is nd + = r bout the -is. Since the upper hlf of this circle is the grph of = f() = r, it follows from (5) tht the volume of the sphere is b r ] r V = π[f()] d = π(r )d = π [r 3 = πr3 r VOLUMES BY WASHERS PERPENDICULAR TO THE -AXIS Not ll solids of revolution hve solid interiors; some hve holes or chnnels tht crete interior surfces, s in Figure 6..8d. So we will lso be interested in problems of the following tpe. r g() f() b 6..5 problem Let f nd g be continuous nd nonnegtive on [,b], nd suppose tht f() g() for ll in the intervl [,b]. Let R be the region tht is bounded bove b = f(), below b = g(), nd on the sides b the lines = nd = b (Figure 6..). Find the volume of the solid of revolution tht is generted b revolving the region R bout the -is (Figure 6..b). Figure 6.. (b) We cn solve this problem b slicing. For this purpose, observe tht the cross section of the solid tken perpendiculr to the -is t the point is the nnulr or wsher-shped

14 46 Chpter 6 / Applictions of the Definite Integrl in Geometr, Science, nd Engineering region with inner rdius g() nd outer rdius f() (Figure 6..b); its re is A() = π[f()] π[g()] = π([f()] [g()] ) Thus, from (3) the volume of the solid is V = b π([f()] [g()] )d (6) Becuse the cross sections re wsher shped, the ppliction of this formul is clled the method of wshers. Emple 4 Find the volume of the solid generted when the region between the grphs of the equtions f() = + nd g() = over the intervl [, ] is revolved bout the -is. Solution. First sketch the region (Figure 6..3); then imgine revolving it bout the -is (Figure 6..3b). From (6) the volume is V = = b π([f()] [g()] )d = ( ) [ π d = π ] π ([ + ] ) d = 69π = + = Unequl scles on es Figure 6..3 Region defined b f nd g () The resulting solid of revolution (b) VOLUMES BY DISKS AND WASHERS PERPENDICULAR TO THE -AXIS The methods of disks nd wshers hve nlogs for regions tht re revolved bout the - is (Figures 6..4 nd 6..5). Using the method of slicing nd Formul (4), ou should be ble to deduce the following formuls for the volumes of the solids in the figures. V = d c Disks π[u()] d V = d c π([w()] [v()] )d Wshers (7 8)

15 6. Volumes b Slicing; Disks nd Wshers 47 d d d = v() d R = u() u() R = w() w() v() c c c c () (b) () (b) Disks Wshers Figure 6..4 Figure 6..5 Emple 5 Find the volume of the solid generted when the region enclosed b =, =, nd = is revolved bout the -is. Solution. First sketch the region nd the solid (Figure 6..6). The cross sections tken perpendiculr to the -is re disks, so we will ppl (7). But first we must rewrite = s =. Thus, from (7) with u() =, the volume is V = d c π[u()] d = π 4 d = π5 5 ] = 3π 5 = = ( = ) Figure 6..6 OTHER AXES OF REVOLUTION It is possible to use the method of disks nd the method of wshers to find the volume of solid of revolution whose is of revolution is line other thn one of the coordinte es. Insted of developing new formul for ech sitution, we will ppel to Formuls (3) nd (4) nd integrte n pproprite cross-sectionl re to find the volume. Emple 6 Find the volume of the solid generted when the region under the curve = over the intervl [, ] is rotted bout the line =. Solution. First sketch the region nd the is of revolution; then imgine revolving the region bout the is (Figure 6..7). At ech in the intervl, the cross section of the solid perpendiculr to the is = is wsher with outer rdius + nd inner rdius. Since the re of this wsher is A() = π([ + ] ) = π( 4 + )

16 48 Chpter 6 / Applictions of the Definite Integrl in Geometr, Science, nd Engineering it follows b (3) tht the volume of the solid is V = A() d = π ( 4 + ) d = π [ ] 3 3 = 76π 5 4 R = Figure 6..7 QUICK CHECK EXERCISES 6. (See pge 43 for nswers.). A solid S etends long the -is from = to = 3. For between nd 3, the cross-sectionl re of S perpendiculr to the -is is 3. An integrl epression for the volume of S is. The vlue of this integrl is.. A solid S is generted b revolving the region between the -is nd the curve = sin ( π) bout the - is. () For between nd π, the cross-sectionl re of S perpendiculr to the -is t is A() =. (b) An integrl epression for the volume of S is. (c) The vlue of the integrl in prt (b) is. 3. A solid S is generted b revolving the region enclosed b the line = + nd the curve = + bout the -is. () For between nd, the crosssectionl re of S perpendiculr to the -is t is A() =. (b) An integrl epression for the volume of S is. 4. A solid S is generted b revolving the region enclosed b the line = + nd the curve = + bout the - is. () For between nd, the crosssectionl re of S perpendiculr to the -is t is A() =. (b) An integrl epression for the volume of S is. EXERCISE SET 6. C CAS 8 Find the volume of the solid tht results when the shded region is revolved bout the indicted is.. = 3 3. = = 3. = 3 4. = /

17 6. Volumes b Slicing; Disks nd Wshers = cos 3 6 = = (, ) = 3 (, 3) = 9. Find the volume of the solid whose bse is the region bounded between the curve = nd the -is from = to = nd whose cross sections tken perpendiculr to the -is re squres.. Find the volume of the solid whose bse is the region bounded between the curve = sec nd the -is from = π/4 to = π/3 nd whose cross sections tken perpendiculr to the -is re squres. 8 Find the volume of the solid tht results when the region enclosed b the given curves is revolved bout the -is.. = 5,= 3. = 9,= 3. =, = /4 4. = sin, = cos, =, = π/4 [Hint: Use the identit cos = cos sin.] 5. = e,=, =, = ln 3 6. = e,=, =, = 7. =,=, =, = 4 + e 3 8. =,=, =, = + e 6 9. Find the volume of the solid whose bse is the region bounded between the curve = 3 nd the -is from = to = nd whose cross sections tken perpendiculr to the -is re squres.. Find the volume of the solid whose bse is the region enclosed between the curve = nd the -is nd whose cross sections tken perpendiculr to the -is re squres. 6 Find the volume of the solid tht results when the region enclosed b the given curves is revolved bout the -is.. = csc, = π/4, = 3π/4, =. =,= 3. =,= + 4. =,= +,=, = 5. = ln, =, =, = 6. = ( > ), =, =, = 7 3 True Flse Determine whether the sttement is true or flse. Eplin our nswer. [In these eercises, ssume tht solid S of volume V is bounded b two prllel plnes perpendiculr to the -is t = nd = b nd tht for ech in [,b], A() denotes the cross-sectionl re of S perpendiculr to the -is.] 7. If ech cross section of S perpendiculr to the -is is squre, then S is rectngulr prllelepiped (i.e., is bo shped). 8. If ech cross section of S is disk or wsher, then S is solid of revolution. 9. If is in centimeters (cm), then A() must be qudrtic function of, since units of A() will be squre centimeters (cm ). 3. The verge vlue of A() on the intervl [,b] is given b V /(b ). 3. Find the volume of the solid tht results when the region bove the -is nd below the ellipse + = ( >,b >) b is revolved bout the -is. 3. Let V be the volume of the solid tht results when the region enclosed b = /, =, =, nd = b( <b<) is revolved bout the -is. Find the vlue of b for which V = Find the volume of the solid generted when the region enclosed b = +,=, nd = is revolved bout the -is. [Hint: Split the solid into two prts.] 34. Find the volume of the solid generted when the region enclosed b =, = 6, nd = is revolved bout the -is. [Hint: Split the solid into two prts.] FOCUS ON CONCEPTS 35. Suppose tht f is continuous function on [,b], nd let R be the region between the curve = f() nd the line = k from = to = b. Using the method of disks, derive with eplntion formul for the volume of solid generted b revolving R bout the line = k. Stte nd eplin dditionl ssumptions, if n, tht ou need bout f for our formul. 36. Suppose tht v nd w re continuous functions on [c, d], nd let R be the region between the curves = v() nd = w() from = c to = d. Using the method of wshers, derive with eplntion formul for the volume of solid generted b revolving R bout the line

18 43 Chpter 6 / Applictions of the Definite Integrl in Geometr, Science, nd Engineering = k. Stte nd eplin dditionl ssumptions, if n, tht ou need bout v nd w for our formul. 37. Consider the solid generted b revolving the shded region in Eercise bout the line =. () Mke conjecture s to which is lrger: the volume of this solid or the volume of the solid in Eercise. Eplin the bsis of our conjecture. (b) Check our conjecture b clculting this volume nd compring it to the volume obtined in Eercise. 38. Consider the solid generted b revolving the shded region in Eercise 4 bout the line =.5. () Mke conjecture s to which is lrger: the volume of this solid or the volume of the solid in Eercise 4. Eplin the bsis of our conjecture. (b) Check our conjecture b clculting this volume nd compring it to the volume obtined in Eercise In prts () (c) find the volume of the solid whose bse is enclosed b the circle + = nd whose cross sections tken perpendiculr to the -is re () semicircles (b) squres (c) equilterl tringles. () (b) 5. As shown in the ccompning figure, cthedrl dome is designed with three semicirculr supports of rdius r so tht ech horizontl cross section is regulr hegon. Show tht the volume of the dome is r 3 3. (c) 39. Find the volume of the solid tht results when the region enclosed b =, =, nd = 9 is revolved bout the line = Find the volume of the solid tht results when the region in Eercise 39 is revolved bout the line = Find the volume of the solid tht results when the region enclosed b = nd = is revolved bout the line =. 4. Find the volume of the solid tht results when the region in Eercise 4 is revolved bout the line =. 43. Find the volume of the solid tht results when the region enclosed b = nd = 3 is revolved bout the line =. 44. Find the volume of the solid tht results when the region in Eercise 43 is revolved bout the line =. 45. A nose cone for spce reentr vehicle is designed so tht cross section, tken ft from the tip nd perpendiculr to the is of smmetr, is circle of rdius 4 ft. Find the volume of the nose cone given tht its length is ft. 46. A certin solid is ft high, nd horizontl cross section tken ft bove the bottom of the solid is n nnulus of inner rdius ft nd outer rdius ft. Find the volume of the solid. 47. Find the volume of the solid whose bse is the region bounded between the curves = nd =, nd whose cross sections perpendiculr to the -is re squres. 48. The bse of certin solid is the region enclosed b =, =, nd = 4. Ever cross section perpendiculr to the -is is semicircle with its dimeter cross the bse. Find the volume of the solid. C r Figure E Use CAS to estimte the volume of the solid tht results when the region enclosed b the curves is revolved bout the stted is. 5. = sin 8, = /π, =, = π/; -is 5. = π sin cos 3, = 4, =, = π/4; -is 53. = e,=, = ; -is 54. = tn, = ; -is 55. The ccompning figure shows sphericl cp of rdius ρ nd height h cut from sphere of rdius r. Show tht the volume V of the sphericl cp cn be epressed s () V = 3 πh (3r h) (b) V = 6 πh(3ρ + h ). h r r Figure E If fluid enters hemisphericl bowl with rdius of ft t rte of ft3 /min, how fst will the fluid be rising when the depth is 5 ft? [Hint: See Eercise 55.] 57. The ccompning figure (on the net pge) shows the dimensions of smll lightbulb t equll spced points. () Use formuls from geometr to mke rough estimte of the volume enclosed b the glss portion of the bulb. (cont.)

19 6. Volumes b Slicing; Disks nd Wshers 43 (b) Use the verge of left nd right endpoint pproimtions to pproimte the volume. r u. cm.45 cm.45 cm Figure E-57. cm.46 cm 5 cm.6 cm.5 cm.5 cm 58. Use the result in Eercise 55 to find the volume of the solid tht remins when hole of rdius r/ is drilled through the center of sphere of rdius r, nd then check our nswer b integrting. 59. As shown in the ccompning figure, cocktil glss with bowl shped like hemisphere of dimeter 8 cm contins cherr with dimeter of cm. If the glss is filled to depth of h cm, wht is the volume of liquid it contins? [Hint: First consider the cse where the cherr is prtill submerged, then the cse where it is totll submerged.] Figure E Find the volume of the torus tht results when the region enclosed b the circle of rdius r with center t (h, ),h>r, is revolved bout the -is. [Hint: Use n pproprite formul from plne geometr to help evlute the definite integrl.] 6. A wedge is cut from right circulr clinder of rdius r b two plnes, one perpendiculr to the is of the clinder nd the other mking n ngle θ with the first. Find the volume of the wedge b slicing perpendiculr to the -is s shown in the ccompning figure..5 cm.5 cm Figure E-6 6. Find the volume of the wedge described in Eercise 6 b slicing perpendiculr to the -is. 63. Two right circulr clinders of rdius r hve es tht intersect t right ngles. Find the volume of the solid common to the two clinders. [Hint: One-eighth of the solid is sketched in the ccompning figure.] 64. In 635 Bonventur Cvlieri, student of Glileo, stted the following result, clled Cvlieri s principle: If two solids hve the sme height, nd if the res of their cross sections tken prllel to nd t equl distnces from their bses re lws equl, then the solids hve the sme volume. Use this result to find the volume of the oblique clinder in the ccompning figure. (See Eercise 5 of Section 6. for plnr version of Cvlieri s principle.) r Figure E-63 r Figure E Writing Use the results of this section to derive Cvlieri s principle (Eercise 64). 66. Writing Write short prgrph tht eplins how Formuls (4) (8) m ll be viewed s consequences of Formul (3). h QUICK CHECK ANSWERS 6.. (b) 3 3 d; 6. () π sin (b) π[ ] d π π sin d (c) π 3. () ; ; π[( + ) ( + ) ]=π[ ] 4. () ; ; π[( ) ( ) ]=π[ + 3 ] (b) π[ + 3 ] d

20 43 Chpter 6 / Applictions of the Definite Integrl in Geometr, Science, nd Engineering 6.3 VOLUMES BY CYLINDRICAL SHELLS The methods for computing volumes tht hve been discussed so fr depend on our bilit to compute the cross-sectionl re of the solid nd to integrte tht re cross the solid. In this section we will develop nother method for finding volumes tht m be pplicble when the cross-sectionl re cnnot be found or the integrtion is too difficult. CYLINDRICAL SHELLS In this section we will be interested in the following problem problem Let f be continuous nd nonnegtive on [,b] ( <b), nd let R be the region tht is bounded bove b = f(), below b the -is, nd on the sides b the lines = nd = b. Find the volume V of the solid of revolution S tht is generted b revolving the region R bout the -is (Figure 6.3.). = f() S R b Figure 6.3. Sometimes problems of the bove tpe cn be solved b the method of disks or wshers perpendiculr to the -is, but when tht method is not pplicble or the resulting integrl is difficult, the method of clindricl shells, which we will discuss here, will often work. A clindricl shell is solid enclosed b two concentric right circulr clinders (Figure 6.3.). The volume V of clindricl shell with inner rdius r, outer rdius r, nd height h cn be written s r r h V =[re of cross section] [height] = (πr πr )h = π(r + r )(r r )h = π [ (r + r ) ] h (r r ) Figure 6.3. But (r + r ) is the verge rdius of the shell nd r r is its thickness, so V = π [verge rdius] [height] [thickness] () We will now show how this formul cn be used to solve Problem The underling ide is to divide the intervl [,b] into n subintervls, thereb subdividing the region R into n strips, R,R,...,R n (Figure 6.3.3). When the region R is revolved bout the -is, these strips generte tube-like solids S,S,...,S n tht re nested one inside the other nd together comprise the entire solid S (Figure 6.3.3b). Thus, the volume V of the solid cn be obtined b dding together the volumes of the tubes; tht is, V = V(S ) + V(S ) + +V(S n )

21 6.3 Volumes b Clindricl Shells 433 = f() S S S 3... S n R R R 3... R n b S Figure () (b) As rule, the tubes will hve curved upper surfces, so there will be no simple formuls for their volumes. However, if the strips re thin, then we cn pproimte ech strip b rectngle (Figure 6.3.4). These rectngles, when revolved bout the -is, will produce clindricl shells whose volumes closel pproimte the volumes of the tubes generted b the originl strips (Figure 6.3.4b). We will show tht b dding the volumes of the clindricl shells we cn obtin Riemnn sum tht pproimtes the volume V, nd b tking the limit of the Riemnn sums we cn obtin n integrl for the ect volume V. R k S k k k Figure Rectngle pproimting the k th strip () Clindricl shell generted b the rectngle (b) Δ k To implement this ide, suppose tht the kth strip etends from k to k nd tht the width of this strip is k = k k If we let k be the midpoint of the intervl [ k, k ], nd if we construct rectngle of height f(k ) over the intervl, then revolving this rectngle bout the -is produces clindricl shell of verge rdius k, height f( k ), nd thickness k (Figure 6.3.5). From (), the volume V k of this clindricl shell is Figure * k k k f(* k ) V k = πk f( k ) k Adding the volumes of the n clindricl shells ields the following Riemnn sum tht pproimtes the volume V : n V πk f( k ) k k= Tking the limit s n increses nd the widths of ll the subintervls pproch zero ields the definite integrl n b V = lim πk m k f( k ) k = πf() d k= In summr, we hve the following result.

22 434 Chpter 6 / Applictions of the Definite Integrl in Geometr, Science, nd Engineering = 6.3. volume b clindricl shells bout the -is Let f be continuous nd nonnegtive on [,b] ( <b), nd let R be the region tht is bounded bove b = f(), below b the -is, nd on the sides b the lines = nd = b. Then the volume V of the solid of revolution tht is generted b revolving the region R bout the -is is given b V = b πf() d () () 4 Emple Use clindricl shells to find the volume of the solid generted when the region enclosed between =, =, = 4, nd the -is is revolved bout the -is. Solution. First sketch the region (Figure 6.3.6); then imgine revolving it bout the -is (Figure 6.3.6b). Since f() =, =, nd b = 4, Formul () ields 4 V = π 4 [ d= π 3/ d = π ] 4 5 5/ = 4π [3 ] =4π 5 5 Cutw view of the solid Figure (b) VARIATIONS OF THE METHOD OF CYLINDRICAL SHELLS The method of clindricl shells is pplicble in vriet of situtions tht do not fit the conditions required b Formul (). For emple, the region m be enclosed between two curves, or the is of revolution m be some line other thn the -is. However, rther thn develop seprte formul for ever possible sitution, we will give generl w of thinking bout the method of clindricl shells tht cn be dpted to ech new sitution s it rises. For this purpose, we will need to reemine the integrnd in Formul (): At ech in the intervl [,b], the verticl line segment from the -is to the curve = f() cn be viewed s the cross section of the region R t (Figure 6.3.7). When the region R is revolved bout the -is, the cross section t sweeps out the surfce of right circulr clinder of height f() nd rdius (Figure 6.3.7b). The re of this surfce is πf() (Figure 6.3.7c), which is the integrnd in (). Thus, Formul () cn be viewed informll in the following w n informl viewpoint bout clindricl shells The volume V of solid of revolution tht is generted b revolving region R bout n is cn be obtined b integrting the re of the surfce generted b n rbitrr cross section of R tken prllel to the is of revolution. = f() c R f () f() b Figure () (b) (c)

23 6.3 Volumes b Clindricl Shells 435 The following emples illustrte how to ppl this result in situtions where Formul () is not pplicble. Emple Use clindricl shells to find the volume of the solid generted when the region R in the first qudrnt enclosed between = nd = is revolved bout the -is (Figure 6.3.8). Solution. As illustrted in prt (b) of Figure 6.3.8, t ech in [, ] the cross section of R prllel to the -is genertes clindricl surfce of height nd rdius. Since the re of this surfce is π( ) the volume of the solid is V = π( )d = π ( 3 )d [ 3 ] [ = π 3 4 = π 4 3 ] = π 4 6 = (, ) = (, ) R = This solid looks like bowl with cone-shped interior. R = Figure () (b) Emple 3 Use clindricl shells to find the volume of the solid generted when the region R under = over the intervl [, ] is revolved bout the line =. Solution. First drw the is of revolution; then imgine revolving the region bout the is (Figure 6.3.9). As illustrted in Figure 6.3.9b, t ech in the intervl 4, the cross section of R prllel to the -is genertes clindricl surfce of height nd rdius +. Since the re of this surfce is π( + )( ) Note tht the volume found in Emple 3 grees with the volume of the sme solid found b the method of wshers in Emple 6 of Section 6.. Confirm tht the volume in Emple found b the method of clindricl shells cn lso be obtined b the method of wshers. it follows tht the volume of the solid is 4 π( + )( )d= π = π 4 ( 3/ + / )d [ 5 5/ + ] 4 3 3/ = 76π 5

24 436 Chpter 6 / Applictions of the Definite Integrl in Geometr, Science, nd Engineering 4 = 4 = R R = = + () (b) Figure QUICK CHECK EXERCISES 6.3 (See pge 438 for nswers.). Let R be the region between the -is nd the curve = + for 4. () For between nd 4, the re of the clindricl surfce generted b revolving the verticl cross section of R t bout the -is is. (b) Using clindricl shells, n integrl epression for the volume of the solid generted b revolving R bout the -is is.. Let R be the region described in Quick Check Eercise. () For between nd 4, the re of the clindricl surfce generted b revolving the verticl cross section of R t bout the line = 5is. (b) Using clindricl shells, n integrl epression for the volume of the solid generted b revolving R bout the line = 5is. 3. A solid S is generted b revolving the region enclosed b the curves = ( ) nd = 4 bout the -is. Using clindricl shells, n integrl epression for the volume of S is. EXERCISE SET 6.3 C CAS 4 Use clindricl shells to find the volume of the solid generted when the shded region is revolved bout the indicted is... = = 4 3. = 4. = + = 4 = 5 Use clindricl shells to find the volume of the solid generted when the region enclosed b the given curves is revolved bout the -is. 5. = 3,=, = 6. =, = 4, = 9, = 7. = /, =, =, = 3 8. = cos( ), =, = π, = 9. =, = + 3, =. =,=. =,=, =, = +. = e,=, = 3,= 3 6 Use clindricl shells to find the volume of the solid generted when the region enclosed b the given curves is revolved bout the -is.

25 6.3 Volumes b Clindricl Shells 437 C C C 3. =, =, = 4. =, =, = 3, = 5. =,=, = 6. = 4, + = 5 7 True Flse Determine whether the sttement is true or flse. Eplin our nswer. 7. The volume of clindricl shell is equl to the product of the thickness of the shell with the surfce re of clinder whose height is tht of the shell nd whose rdius is equl to the verge of the inner nd outer rdii of the shell. 8. The method of clindricl shells is specil cse of the method of integrtion of cross-sectionl re tht ws discussed in Section In the method of clindricl shells, integrtion is over n intervl on coordinte is tht is perpendiculr to the is of revolution of the solid.. The Riemnn sum pproimtion.. 3. V n πk f( k ) k k= ( where k = ) k + k for the volume of solid of revolution is ect when f is constnt function. Use CAS to find the volume of the solid generted when the region enclosed b = e nd = for is revolved bout the -is. Use CAS to find the volume of the solid generted when the region enclosed b = cos, =, nd = for π/ is revolved bout the -is. Consider the region to the right of the -is, to the left of the verticl line = k( <k<π), nd between the curve = sin nd the -is. Use CAS to estimte the vlue of k so tht the solid generted b revolving the region bout the -is hs volume of 8 cubic units. FOCUS ON CONCEPTS 4. Let R nd R be regions of the form shown in the ccompning figure. Use clindricl shells to find formul for the volume of the solid tht results when () region R is revolved bout the -is (b) region R is revolved bout the -is. = f() R = g() Figure E-4 b d c = g() R = f() 5. () Use clindricl shells to find the volume of the solid tht is generted when the region under the curve = over [, ] is revolved bout the -is. (b) For this problem, is the method of clindricl shells esier or hrder thn the method of slicing discussed in the lst section? Eplin. 6. Let f be continuous nd nonnegtive on [,b], nd let R be the region tht is enclosed b = f() nd = for b. Using the method of clindricl shells, derive with eplntion formul for the volume of the solid generted b revolving R bout the line = k, where k. 7 8 Using the method of clindricl shells, set up but do not evlute n integrl for the volume of the solid generted when the region R is revolved bout () the line = nd (b) the line =. 7. R is the region bounded b the grphs of =, =, nd =. 8. R is the region in the first qudrnt bounded b the grphs of =, =, nd =. 9. Use clindricl shells to find the volume of the solid tht is generted when the region tht is enclosed b = / 3, =, =, = is revolved bout the line =. 3. Use clindricl shells to find the volume of the solid tht is generted when the region tht is enclosed b = 3, =, = is revolved bout the line =. 3. Use clindricl shells to find the volume of the cone generted when the tringle with vertices (, ), (, r), (h, ), where r> nd h>, is revolved bout the -is. 3. The region enclosed between the curve = k nd the line = 4 k is revolved bout the line = k. Use clindricl shells to find the volume of the resulting solid. (Assume k>.) 33. As shown in the ccompning figure, clindricl hole is drilled ll the w through the center of sphere. Show tht the volume of the remining solid depends onl on the length L of the hole, not on the size of the sphere. L Figure E Use clindricl shells to find the volume of the torus obtined b revolving the circle + = bout the line

26 438 Chpter 6 / Applictions of the Definite Integrl in Geometr, Science, nd Engineering = b, where b>>. [Hint: It m help in the integrtion to think of n integrl s n re.] 35. Let V nd V be the volumes of the solids tht result when the region enclosed b = /, =, =, nd = b ( b> ) is revolved bout the -is nd -is, respectivel. Is there vlue of b for which V = V? 36. () Find the volume V of the solid generted when the region bounded b = /( + 4 ), =, =, nd = b(b>) is revolved bout the -is. (b) Find lim b + V. 37. Writing Fced with the problem of computing the volume of solid of revolution, how would ou go bout deciding whether to use the method of disks/wshers or the method of clindricl shells? 38. Writing With both the method of disks/wshers nd with the method of clindricl shells, we integrte n re to get the volume of solid of revolution. However, these two pproches differ in ver significnt ws. Write brief prgrph tht discusses these differences. QUICK CHECK ANSWERS 6.3. () π( + ) (b) π[4 ( ) ] d π( + )d. () π(5 )( + ) (b) 4 π(5 )( + )d 6.4 LENGTH OF A PLANE CURVE In this section we will use the tools of clculus to stud the problem of finding the length of plne curve. = f () b ARC LENGTH Our first objective is to define wht we men b the length (lso clled the rc length) of plne curve = f() over n intervl [,b] (Figure 6.4.). Once tht is done we will be ble to focus on the problem of computing rc lengths. To void some complictions tht would otherwise occur, we will impose the requirement tht f be continuous on [,b],in which cse we will s tht = f() is smooth curve on [,b] or tht f is smooth function on [,b]. Thus, we will be concerned with the following problem. Figure 6.4. Intuitivel, ou might think of the rc length of curve s the number obtined b ligning piece of string with the curve nd then mesuring the length of the string fter it is strightened out rc length problem Suppose tht = f() is smooth curve on the intervl [,b]. Define nd find formul for the rc length L of the curve = f() over the intervl [,b]. To define the rc length of curve we strt b breking the curve into smll segments. Then we pproimte the curve segments b line segments nd dd the lengths of the line segments to form Riemnn sum. Figure 6.4. illustrtes how such line segments tend to become better nd better pproimtions to curve s the number of segments increses. As the number of segments increses, the corresponding Riemnn sums pproch definite integrl whose vlue we will tke to be the rc length L of the curve. To implement our ide for solving Problem 6.4., divide the intervl [,b] into n subintervls b inserting points,,..., n between = nd b = n. As shown in Figure 6.4.3, let P,P,...,P n be the points on the curve with -coordintes =,

27 6.4 Length of Plne Curve 439 Figure 6.4. Shorter line segments provide better pproimtion to the curve. P n P P P P 3 f ( k ) f ( k ) L k P k Δ k P k Δ k 3... n k k = b = n Figure () (b),,..., n,b = n nd join these points with stright line segments. These line segments form polgonl pth tht we cn regrd s n pproimtion to the curve = f(). As indicted in Figure 6.4.3b, the length L k of the kth line segment in the polgonl pth is L k = ( k ) + ( k ) = ( k ) +[f( k ) f( k )] () If we now dd the lengths of these line segments, we obtin the following pproimtion to the length L of the curve Eplin wh the pproimtion in () cnnot be greter thn L. L n L k = k= n ( k ) +[f( k ) f( k )] () k= To put this in the form of Riemnn sum we will ppl the Men-Vlue Theorem (4.8.). This theorem implies tht there is point k between k nd k such tht f( k ) f( k ) k k = f ( k ) or f( k) f( k ) = f ( k ) k nd hence we cn rewrite () s n n L ( k ) +[f (k )] ( k ) = +[f (k )] k k= Thus, tking the limit s n increses nd the widths of ll the subintervls pproch zero ields the following integrl tht defines the rc length L: n b L = lim +[f (k m k )] k = +[f ()] d k= In summr, we hve the following definition. k=

28 44 Chpter 6 / Applictions of the Definite Integrl in Geometr, Science, nd Engineering 6.4. definition If = f() is smooth curve on the intervl [,b], then the rc length L of this curve over [,b] is defined s L = b +[f ()] d (3) This result provides both definition nd formul for computing rc lengths. Where convenient, (3) cn lso be epressed s L = b +[f ()] d = b + ( ) d d (4) d Moreover, for curve epressed in the form = g(), where g is continuous on [c, d], the rc length L from = c to = d cn be epressed s L = d c +[g ()] d = d c + ( ) d d (5) d Figure (, ) (, ) = 3/ Emple Find the rc length of the curve = 3/ from (, ) to (, ) (Figure 6.4.4) in two ws: () using Formul (4) nd (b) using Formul (5). Solution (). d d = 3 / nd since the curve etends from = to =, it follows from (4) tht L = + ( 3 /) d = d To evlute this integrl we mke the u-substitution u = + 9 4, du = 9 4 d nd ( then chnge the -limits of integrtion ( =, = ) to the corresponding u-limits u = 3 4,u= ) 4 : L = 4 9 /4 3/4 u / du = 8 ] /4 7 u3/ = 8 3/4 7 [ ( ) 3/ 4 ( ) ] 3 3/ 4 = Solution (b). To ppl Formul (5) we must first rewrite the eqution = 3/ so tht is epressed s function of. This ields = /3 nd d d = 3 /3 Since the curve etends from = to =, it follows from (5) tht L = /3 d = 3 /3 9 /3 + 4 d

29 6.4 Length of Plne Curve 44 The rc from the point (, ) to the point (, ) in Figure is nerl stright line, so the rc length should be onl slightl lrger thn the strightline distnce between these points. Show tht this is so. To evlute this integrl we mke the u-substitution 3 u = 9 /3 + 4, 3 du = 6 /3 d nd chnge the -limits of integrtion ( =, = ) to the corresponding u-limits (u = 3,u= ). This gives L = u / du = ] 8 7 u3/ = 7 [()3/ (3) 3/ ]= The nswer in prt (b) grees with tht in prt (); however, the integrtion in prt (b) is more tedious. In problems where there is choice between using (4) or (5), it is often the cse tht one of the formuls leds to simpler integrl thn the other. FINDING ARC LENGTH BY NUMERICAL METHODS In the net chpter we will develop some techniques of integrtion tht will enble us to find ect vlues of more integrls encountered in rc length clcultions; however, generll speking, most such integrls re impossible to evlute in terms of elementr functions. In these cses one usull pproimtes the integrl using numericl method such s the midpoint rule discussed in Section 5.4. TECHNOLOGY MASTERY If our clculting utilit hs numericl integrtion cpbilit, use it to confirm tht the rc length L in Emple is pproimtel L 3.8. Emple From (4), the rc length of = sin from = to = π is given b the integrl π L = + (cos ) d This integrl cnnot be evluted in terms of elementr functions; however, using clculting utilit with numericl integrtion cpbilit ields the pproimtion L 3.8. QUICK CHECK EXERCISES 6.4 (See pge 443 for nswers.). A function f is smooth on [,b] if f is on [,b].. If function f is smooth on [,b], then the length of the curve = f() over [,b] is. 3. The distnce between points (, ) nd (e, ) is. 4. Let L be the length of the curve = ln from (, ) to (e, ). () Integrting with respect to, n integrl epression for L is. (b) Integrting with respect to, n integrl epression for L is. EXERCISE SET 6.4 C CAS. Use the Theorem of Pthgors to find the length of the line segment = from (, ) to (, 4), nd confirm tht the vlue is consistent with the length computed using () Formul (4) (b) Formul (5).. Use the Theorem of Pthgors to find the length of the line segment = 5 from (, ) nd (, 5), nd confirm tht the vlue is consistent with the length computed using () Formul (4) (b) Formul (5). 3 8 Find the ect rc length of the curve over the intervl. 3. = 3 3/ from = to = 4. = 3 ( + ) 3/ from = to = 5. = /3 from = to = 8 6. = ( 6 + 8)/(6 ) from = to = = from = to = 4 8. = from = to = 4 9 True Flse Determine whether the sttement is true or flse. Eplin our nswer. 9. The grph of = is smooth curve on [, ].

30 44 Chpter 6 / Applictions of the Definite Integrl in Geometr, Science, nd Engineering C. The pproimtion n L ( k ) +[f( k ) f( k )] k= for rc length is not epressed in the form of Riemnn sum.. The pproimtion n L +[f (k )] k k= for rc length is ect when f is liner function of.. In our definition of the rc length for the grph of = f(), we need f () to be continuous function in order for f to stisf the hpotheses of the Men-Vlue Theorem (4.8.). 3 4 Epress the ect rc length of the curve over the given intervl s n integrl tht hs been simplified to eliminte the rdicl, nd then evlute the integrl using CAS. 3. = ln(sec ) from = to = π/4 4. = ln(sin ) from = π/4 to = π/ (g) Use clculting utilit with numericl integrtion cpbilities to pproimte the rc length integrls in prt (b) to four deciml plces. 8. Follow the directions of Eercise 7 for the curve segments = 8/3 from = 3 to = nd = 3/8 from = 8 to =. 9. Follow the directions of Eercise 7 for the curve segment = tn from = to = π/3 nd for the curve segment = tn from = to = 3.. Let = f() be smooth curve on the closed intervl [,b]. Prove tht if m nd M re nonnegtive numbers such tht m f () M for ll in [,b], then the rc length L of = f()over the intervl [,b] stisfies the inequlities (b ) + m L (b ) + M. Use the result of Eercise to show tht the rc length L of = sec over the intervl π/3 stisfies π 3 L π 3 3 FOCUS ON CONCEPTS 5. Consider the curve = /3. () Sketch the portion of the curve between = nd = 8. (b) Eplin wh Formul (4) cnnot be used to find the rc length of the curve sketched in prt (). (c) Find the rc length of the curve sketched in prt (). 6. The curve segment = from = to = m lso be epressed s the grph of = from = to = 4. Set up two integrls tht give the rc length of this curve segment, one b integrting with respect to, nd the other b integrting with respect to. Demonstrte substitution tht verifies tht these two integrls re equl. 7. Consider the curve segments = from = to = nd = from = to = 4. 4 () Grph the two curve segments nd use our grphs to eplin wh the lengths of these two curve segments should be equl. (b) Set up integrls tht give the rc lengths of the curve segments b integrting with respect to. Demonstrte substitution tht verifies tht these two integrls re equl. (c) Set up integrls tht give the rc lengths of the curve segments b integrting with respect to. (d) Approimte the rc length of ech curve segment using Formul () with n = equl subintervls. (e) Which of the two pproimtions in prt (d) is more ccurte? Eplin. (f ) Use the midpoint pproimtion with n = subintervls to pproimte ech rc length integrl in prt (b). C C. 3. A bsketbll pler mkes successful shot from the free throw line. Suppose tht the pth of the bll from the moment of relese to the moment it enters the hoop is described b = , 4.6 where is the horizontl distnce (in meters) from the point of relese, nd is the verticl distnce (in meters) bove the floor. Use CAS or scientific clcultor with numericl integrtion cpbilit to pproimte the distnce the bll trvels from the moment it is relesed to the moment it enters the hoop. Round our nswer to two deciml plces. The centrl spn of the Golden Gte Bridge in Cliforni is 4 ft long nd is suspended from cbles tht rise 5 ft bove the rodw on either side. Approimtel how long is the portion of cble tht lies between the support towers on one side of the rodw? [Hint: As suggested b the ccompning figure, ssume the cble is modeled b prbol = tht psses through the point (, 5). Use CAS or clculting utilit with numericl integrtion cpbilit to pproimte the length of the cble. Round our nswer to the nerest foot.] Figure E-3 (, 5)

31 6.4 Length of Plne Curve 443 C C C As shown in the ccompning figure, horizontl bem with dimensions in 6in 6 ft is fied t both ends nd is subjected to uniforml distributed lod of lb/ft. As result of the lod, the centerline of the bem undergoes deflection tht is described b =.67 8 ( 4 L 3 + L ) ( 9), where L = 9 in is the length of the unloded bem, is the horizontl distnce long the bem mesured in inches from the left end, nd is the deflection of the centerline in inches. () Grph versus for 9. (b) Find the mimum deflection of the centerline. (c) Use CAS or clcultor with numericl integrtion cpbilit to find the length of the centerline of the loded bem. Round our nswer to two deciml plces. = = 9 Figure E-4 A golfer mkes successful chip shot to the green. Suppose tht the pth of the bll from the moment it is struck to the moment it hits the green is described b =.54.4 where is the horizontl distnce (in rds) from the point where the bll is struck, nd is the verticl distnce (in rds) bove the firw. Use CAS or clculting utilit with numericl integrtion cpbilit to find the distnce the bll trvels from the moment it is struck to the moment it hits the green. Assume tht the firw nd green re t the sme level nd round our nswer to two deciml plces These eercises ssume fmilirit with the bsic concepts of prmetric curves. If needed, n introduction to this mteril is provided in Web Appendi I. 6. Assume tht no segment of the curve = (t), = (t), ( t b) is trced more thn once s t increses from to b. Divide the intervl [,b] into n subintervls b inserting points t,t,...,t n between = t nd b = t n. Let L denote the rc length of the curve. Give n informl rgument for the pproimtion n L [(tk ) (t k )] +[(t k ) (t k )] k= C If d/dt nd d/dt re continuous functions for t b, then it cn be shown tht s m t k, this sum converges to L = b ( ) d + dt ( ) d dt dt 7 3 Use the rc length formul from Eercise 6 to find the rc length of the curve. 7. = 3 t 3,= t ( t ) 8. = ( + t),=( + t) 3 ( t ) 9. = cos t, = sin t ( t π/) 3. = cos t + t sin t, = sin t t cos t ( t π) 3. = e t cos t, = e t sin t ( t π/) 3. = e t (sin t + cos t), = e t (cos t sin t) ( t 4) 33. () Show tht the totl rc length of the ellipse = cos t, = sin t ( t π) is given b 4 π/ + 3 sin tdt (b) Use CAS or scientific clcultor with numericl integrtion cpbilit to pproimte the rc length in prt (). Round our nswer to two deciml plces. (c) Suppose tht the prmetric equtions in prt () describe the pth of prticle moving in the -plne, where t is time in seconds nd nd re in centimeters. Use CAS or scientific clcultor with numericl integrtion cpbilit to pproimte the distnce trveled b the prticle from t =.5stot = 4.8 s. Round our nswer to two deciml plces. 34. Show tht the totl rc length of the ellipse = cos t, = b sin t, t π for >b>is given b π/ 4 k cos tdt where k = b /. 35. Writing In our discussion of Arc Length Problem 6.4., we derived the pproimtion n L +[f (k )] k k= Discuss the geometric mening of this pproimtion. (Be sure to ddress the ppernce of the derivtive f.) 36. Writing Give emples in which Formul (4) for rc length cnnot be pplied directl, nd describe how ou would go bout finding the rc length of the curve in ech cse. (Discuss both the use of lterntive formuls nd the use of numericl methods.) QUICK CHECK ANSWERS 6.4. continuous. b +[f ()] d 3. (e ) + 4. () e + ( /) d (b) + e d

32 444 Chpter 6 / Applictions of the Definite Integrl in Geometr, Science, nd Engineering 6.5 AREA OF A SURFACE OF REVOLUTION In this section we will consider the problem of finding the re of surfce tht is generted b revolving plne curve bout line. SURFACE AREA A surfce of revolution is surfce tht is generted b revolving plne curve bout n is tht lies in the sme plne s the curve. For emple, the surfce of sphere cn be generted b revolving semicircle bout its dimeter, nd the lterl surfce of right circulr clinder cn be generted b revolving line segment bout n is tht is prllel to it (Figure 6.5.). Some Surfces of Revolution Figure 6.5. In this section we will be concerned with the following problem. = f() 6.5. surfce re problem Suppose tht f is smooth, nonnegtive function on [,b] nd tht surfce of revolution is generted b revolving the portion of the curve = f() between = nd = b bout the -is (Figure 6.5.). Define wht is ment b the re S of the surfce, nd find formul for computing it. S Figure 6.5. b b To motivte n pproprite definition for the re S of surfce of revolution, we will decompose the surfce into smll sections whose res cn be pproimted b elementr formuls, dd the pproimtions of the res of the sections to form Riemnn sum tht pproimtes S, nd then tke the limit of the Riemnn sums to obtin n integrl for the ect vlue of S. To implement this ide, divide the intervl [,b] into n subintervls b inserting points,,..., n between = nd b = n. As illustrted in Figure 6.5.3, the corresponding points on the grph of f define polgonl pth tht pproimtes the curve = f()over the intervl [,b]. As illustrted in Figure 6.5.3b, when this polgonl pth is revolved bout the -is, it genertes surfce consisting of n prts, ech of which is portion of right circulr cone clled frustum (from the Ltin mening bit or piece ). Thus, the re of ech prt of the pproimting surfce cn be obtined from the formul S = π(r + r )l () for the lterl re S of frustum of slnt height l nd bse rdii r nd r (Figure 6.5.4). As suggested b Figure 6.5.5, the kth frustum hs rdii f( k ) nd f( k ) nd height k. Its slnt height is the length L k of the kth line segment in the polgonl pth, which from Formul () of Section 6.4 is L k = ( k ) +[f( k ) f( k )]

33 6.5 Are of Surfce of Revolution 445 = f () r l =... n b = n r Figure () (b) Figure Frustum Δ k f( k ) k k L k Figure f( k ) This mkes the lterl re S k of the kth frustum S k = π[f( k ) + f( k )] ( k ) +[f( k ) f( k )] If we dd these res, we obtin the following pproimtion to the re S of the entire surfce: n S π[f( k ) + f( k )] ( k ) +[f( k ) f( k )] () k= To put this in the form of Riemnn sum we will ppl the Men-Vlue Theorem (4.8.). This theorem implies tht there is point k between k nd k such tht f( k ) f( k ) = f (k k ) k or f( k) f( k ) = f (k ) k nd hence we cn rewrite () s n S π[f( k ) + f( k )] ( k ) +[f (k )] ( k ) = k= n k= π[f( k ) + f( k )] +[f (k )] k (3) However, this is not et Riemnn sum becuse it involves the vribles k nd k. To eliminte these vribles from the epression, observe tht the verge vlue of the numbers f( k ) nd f( k ) lies between these numbers, so the continuit of f nd the Intermedite-Vlue Theorem (.5.7) impl tht there is point k between k nd k such tht [f( k ) + f( k )]=f(k ) Thus, () cn be epressed s S n k= πf( k ) +[f ( k )] k Although this epression is close to Riemnn sum in form, it is not true Riemnn sum becuse it involves two vribles k nd k, rther thn k lone. However, it is proved in dvnced clculus courses tht this hs no effect on the limit becuse of the continuit of f. Thus, we cn ssume tht k = k when tking the limit, nd this suggests tht S cn be defined s n b S = lim πf(k ) +[f (k m k )] k = πf() +[f ()] d k= In summr, we hve the following definition.

34 446 Chpter 6 / Applictions of the Definite Integrl in Geometr, Science, nd Engineering 6.5. definition If f is smooth, nonnegtive function on [,b], then the surfce re S of the surfce of revolution tht is generted b revolving the portion of the curve = f() between = nd = b bout the -is is defined s S = b πf() +[f ()] d This result provides both definition nd formul for computing surfce res. Where convenient, this formul cn lso be epressed s S = b πf() +[f ()] d = b π + ( ) d d (4) d Moreover, if g is nonnegtive nd = g() is smooth curve on the intervl [c, d], then the re of the surfce tht is generted b revolving the portion of curve = g() between = c nd = d bout the -is cn be epressed s S = d c πg() +[g ()] d = d c π + ( ) d d (5) d = 3 (, ) Emple Find the re of the surfce tht is generted b revolving the portion of the curve = 3 between = nd = bout the -is. Solution. First sketch the curve; then imgine revolving it bout the -is (Figure 6.5.6). Since = 3,wehved/d = 3, nd hence from (4) the surfce re S is ( ) d S = π + d d = π 3 + (3 ) d Figure = π 3 ( ) / d = π 36 u / du u = du = 36 3 d (, 4) = π 36 ] 3 u3/ = π u= 7 (3/ ) 3.56 Figure (, ) = Emple Find the re of the surfce tht is generted b revolving the portion of the curve = between = nd = bout the -is. Solution. First sketch the curve; then imgine revolving it bout the -is (Figure 6.5.7). Becuse the curve is revolved bout the -is we will ppl Formul (5). Towrd this end, we rewrite = s = nd observe tht the -vlues corresponding to = nd

35 6.5 Are of Surfce of Revolution 447 = re = nd = 4. Since =,wehved/d = /( ), nd hence from (5) the surfce re S is 4 ( ) d S = π + d d 4 = π ( ) + d = π = π = π 4 3 u3/ 4 + d u / du ] 7 u=5 u = 4 + du = 4 d = π 6 (73/ 5 3/ ) 3.85 QUICK CHECK EXERCISES 6.5 (See pge 449 for nswers.). If f is smooth, nonnegtive function on [,b], then the surfce re S of the surfce of revolution generted b revolving the portion of the curve = f() between = nd = b bout the -is is.. The lterl re of the frustum with slnt height nd bse rdii r = nd r = is. 3. An integrl epression for the re of the surfce generted b rotting the line segment joining (3, ) nd (6, ) bout the -is is. 4. An integrl epression for the re of the surfce generted b rotting the line segment joining (3, ) nd (6, ) bout the -is is. EXERCISE SET 6.5 C CAS C 4 Find the re of the surfce generted b revolving the given curve bout the -is.. = 7,. =, 4 3. = 4, 4. = 3, Find the re of the surfce generted b revolving the given curve bout the -is. 5. = 9 +, 6. = 3, 7. = 9, 8. =, 9 Use CAS to find the ect re of the surfce generted b revolving the curve bout the stted is. 9. = 3 3/, 3; -is. = , ; -is C. 8 = 6 +, ; -is. = 6, 5; -is 3 6 Use CAS or clculting utilit with numericl integrtion cpbilit to pproimte the re of the surfce generted b revolving the curve bout the stted is. Round our nswer to two deciml plces. 3. = sin, π; -is 4. = tn, π/4; -is 5. = e, ; -is 6. = e, e; -is 7 True Flse Determine whether the sttement is true or flse. Eplin our nswer. 7. The lterl surfce re S of right circulr cone with height h nd bse rdius r is S = πr r + h. 8. The lterl surfce re of frustum of slnt height l nd bse rdii r nd r is equl to the lterl surfce re of right circulr clinder of height l nd rdius equl to the verge of r nd r.

36 448 Chpter 6 / Applictions of the Definite Integrl in Geometr, Science, nd Engineering 9. The pproimtion n S πf (k ) +[f (k )] k k= for surfce re is ect if f is positive-vlued constnt function.. The epression n k= is not true Riemnn sum for b πf ( k ) +[f ( k )] k πf () +[f ()] d Approimte the re of the surfce using Formul () with n = subintervls of equl width. Round our nswer to two deciml plces.. The surfce of Eercise 3.. The surfce of Eercise 6. FOCUS ON CONCEPTS 3. Assume tht = f() is smooth curve on the intervl [,b] nd ssume tht f() for b. Derive formul for the surfce re generted when the curve = f(), b, is revolved bout the line = k(k>). 4. Would it be circulr resoning to use Definition 6.5. to find the surfce re of frustum of right circulr cone? Eplin our nswer. 5. Show tht the re of the surfce of sphere of rdius r is 4πr. [Hint: Revolve the semicircle = r bout the -is.] 6. The ccompning figure shows sphericl cp of height h cut from sphere of rdius r. Show tht the surfce re S of the cp is S = πrh. [Hint: Revolve n pproprite portion of the circle + = r bout the -is.] h r Figure E-6 7. The portion of sphere tht is cut b two prllel plnes is clled zone. Use the result of Eercise 6 to show tht the surfce re of zone depends on the rdius of the sphere nd the distnce between the plnes, but not on the loction of the zone. 8. Let = f() be smooth curve on the intervl [,b] nd ssume tht f() for b. B the Etreme-Vlue Theorem (4.4.), the function f hs mimum vlue K nd minimum vlue k on [,b]. Prove: If L is the rc length of the curve = f() between = nd = b, nd if S is the re of the surfce tht is generted b revolving this curve bout the -is, then πkl S πkl 9. Use the results of Eercise 8 bove nd Eercise in Section 6.4 to show tht the re S of the surfce generted b revolving the curve = sec, π/3, bout the -is stisfies π 3 S 4π Let = f() be smooth curve on [,b] nd ssume tht f() for b. Let A be the re under the curve = f()between = nd = b, nd let S be the re of the surfce obtined when this section of curve is revolved bout the -is. () Prove tht πa S. (b) For wht functions f is πa = S? 3 37 These eercises ssume fmilirit with the bsic concepts of prmetric curves. If needed, n introduction to this mteril is provided in Web Appendi I. 3 3 For these eercises, divide the intervl [,b] into n subintervls b inserting points t,t,...,t n between = t nd b = t n, nd ssume tht (t) nd (t) re continuous functions nd tht no segment of the curve = (t), = (t) ( t b) is trced more thn once. 3. Let S be the re of the surfce generted b revolving the curve = (t), = (t) ( t b) bout the -is. Eplin how S cn be pproimted b n S (π[(t k ) + (t k )] k= [(t k ) (t k )] +[(t k ) (t k )] ) Using results from dvnced clculus, it cn be shown tht s m t k, this sum converges to S = b π(t) [ (t)] +[ (t)] dt (A) 3. Let S be the re of the surfce generted b revolving the curve = (t), = (t) ( t b) bout the -is. Eplin how S cn be pproimted b n S (π[(t k ) + (t k )] k= [(t k ) (t k )] +[(t k ) (t k )] ) Using results from dvnced clculus, it cn be shown tht s m t k, this sum converges to S = b π(t) [ (t)] +[ (t)] dt (B)

37 6.6 Work 449 C Use Formuls (A) nd (B) from Eercises 3 nd Find the re of the surfce generted b revolving the prmetric curve = t, = t ( t 4) bout the -is. 34. Use CAS to find the re of the surfce generted b revolving the prmetric curve = cos t, = 5 sin t ( t π/) bout the -is. 35. Find the re of the surfce generted b revolving the prmetric curve = t, = t ( t ) bout the -is. 36. Find the re of the surfce generted b revolving the prmetric curve = cos t, = sin t( t π/) bout the -is. 37. B revolving the semicircle = r cos t, = r sin t ( t π) bout the -is, show tht the surfce re of sphere of rdius r is 4πr. 38. Writing Compre the derivtion of Definition 6.5. with tht of Definition Discuss the geometric fetures tht result in similrities in the two definitions. 39. Writing Discuss wht goes wrong if we replce the frustums of right circulr cones b right circulr clinders in the derivtion of Definition QUICK CHECK ANSWERS 6.5. b πf() +[f ()] d. 3 π ( (π) 3 ) 6 9 d = 3 π d 4. 9 (π)(3) d 6.6 WORK In this section we will use the integrtion tools developed in the preceding chpter to stud some of the bsic principles of work, which is one of the fundmentl concepts in phsics nd engineering. THE ROLE OF WORK IN PHYSICS AND ENGINEERING In this section we will be concerned with two relted concepts, work nd energ. To put these ides in fmilir setting, when ou push stlled cr for certin distnce ou re performing work, nd the effect of our work is to mke the cr move. The energ of motion cused b the work is clled the kinetic energ of the cr. The ect connection between work nd kinetic energ is governed b principle of phsics clled the work energ reltionship. Although we will touch on this ide in this section, detiled stud of the reltionship between work nd energ will be left for courses in phsics nd engineering. Our primr gol here will be to eplin the role of integrtion in the stud of work. WORK DONE BY A CONSTANT FORCE APPLIED IN THE DIRECTION OF MOTION When stlled cr is pushed, the speed tht the cr ttins depends on the force F with which it is pushed nd the distnce d over which tht force is pplied (Figure 6.6.). Force nd distnce pper in the following definition of work. d F F Figure 6.6.

38 45 Chpter 6 / Applictions of the Definite Integrl in Geometr, Science, nd Engineering 6.6. definition If constnt force of mgnitude F is pplied in the direction of motion of n object, nd if tht object moves distnce d, then we define the work W performed b the force on the object to be W = F d () If ou push ginst n immovble object, such s brick wll, ou m tire ourself out, but ou will not perform n work. Wh? Common units for mesuring force re newtons (N) in the Interntionl Sstem of Units (SI), dnes (dn) in the centimeter-grm-second (CGS) sstem, nd pounds (lb) in the British Engineering (BE) sstem. One newton is the force required to give mss of kg n ccelertion of m/s, one dne is the force required to give mss ofgnccelertion of cm/s, nd one pound of force is the force required to give mss of slug n ccelertion of ft/s. It follows from Definition 6.6. tht work hs units of force times distnce. The most common units of work re newton-meters (N m), dne-centimeters (dn cm), nd footpounds (ft lb). As indicted in Tble 6.6., one newton-meter is lso clled joule (J), nd one dne-centimeter is lso clled n erg. One foot-pound is pproimtel.36 J. Tble 6.6. sstem force distnce = work SI CGS BE newton (N) dne (dn) pound (lb) meter (m) centimeter (cm) foot (ft) joule (J) erg foot-pound (ft lb) conversion fctors: N = 5 dn.5 lb lb 4.45 N J = 7 erg.738 ft lb ft lb.36 J =.36 7 erg Emple An object moves 5 ft long line while subjected to constnt force of lb in its direction of motion. The work done is W = F d = 5 = 5 ft lb An object moves 5 m long line while subjected to constnt force of 4 N in its direction of motion. The work done is W = F d = 4 5 = N m = J Emple In the 976 Olmpics, Vsili Aleeev stounded the world b lifting record-breking 56 lb from the floor to bove his hed (bout m). Equll stounding ws the fet of strongmn Pul Anderson, who in 957 brced himself on the floor nd used his bck to lift 67 lb of led nd utomobile prts distnce of cm. Who did more work? Stephen Sutton/DUOMO Archive/PCN Photogrph Inc. Vsili Aleeev shown lifting recordbreking 56 lb in the 976 Olmpics. In eight successive ers he won Olmpic gold medls, cptured si world chmpionships, nd broke 8 world records. In 999 he ws honored in Greece s the best sportsmn of the twentieth centur. Solution. To lift n object one must ppl sufficient force to overcome the grvittionl force tht the Erth eerts on tht object. The force tht the Erth eerts on n object is tht object s weight; thus, in performing their fets, Aleeev pplied force of 56 lb over distnce of m nd Anderson pplied force of 67 lb over distnce of cm. Pounds re units in the BE sstem, meters re units in SI, nd centimeters re units in the CGS sstem. We will need to decide on the mesurement sstem we wnt to use nd be consistent. Let us gree to use SI nd epress the work of the two men in joules. Using the conversion fctor in Tble 6.6. we obtin 56 lb 56 lb 4.45 N/lb 5 N 67 lb 67 lb 4.45 N/lb 7,9 N

39 6.6 Work 45 Using these vlues nd the fct tht cm =. m we obtin Aleeev s work = (5 N) ( m) = 5 J Anderson s work = (7,9 N) (. m) = 79 J Therefore, even though Anderson s lift required tremendous upwrd force, it ws pplied over such short distnce tht Aleeev did more work. () Nturl position Force must be eerted to stretch spring WORK DONE BY A VARIABLE FORCE APPLIED IN THE DIRECTION OF MOTION Mn importnt problems re concerned with finding the work done b vrible force tht is pplied in the direction of motion. For emple, Figure 6.6. shows spring in its nturl stte (neither compressed nor stretched). If we wnt to pull the block horizontll (Figure 6.6.b), then we would hve to ppl more nd more force to the block to overcome the incresing force of the stretching spring. Thus, our net objective is to define wht is ment b the work performed b vrible force nd to find formul for computing it. This will require clculus. Figure 6.6. (b) 6.6. problem Suppose tht n object moves in the positive direction long coordinte line while subjected to vrible force F()tht is pplied in the direction of motion. Define wht is ment b the work W performed b the force on the object s the object moves from = to = b, nd find formul for computing the work. The bsic ide for solving this problem is to brek up the intervl [,b] into subintervls tht re sufficientl smll tht the force does not vr much on ech subintervl. This will llow us to tret the force s constnt on ech subintervl nd to pproimte the work on ech subintervl using Formul (). B dding the pproimtions to the work on the subintervls, we will obtin Riemnn sum tht pproimtes the work W over the entire intervl, nd b tking the limit of the Riemnn sums we will obtin n integrl for W. To implement this ide, divide the intervl [,b] into n subintervls b inserting points,,..., n between = nd b = n. We cn use Formul () to pproimte the work W k done in the kth subintervl b choosing n point k in this intervl nd regrding the force to hve constnt vlue F(k ) throughout the intervl. Since the width of the kth subintervl is k k = k, this ields the pproimtion W k F( k ) k Adding these pproimtions ields the following Riemnn sum tht pproimtes the work W done over the entire intervl: n W F(k ) k k= Tking the limit s n increses nd the widths of ll the subintervls pproch zero ields the definite integrl W = lim m k n F(k ) k = k= In summr, we hve the following result. b F()d

40 45 Chpter 6 / Applictions of the Definite Integrl in Geometr, Science, nd Engineering definition Suppose tht n object moves in the positive direction long coordinte line over the intervl [,b] while subjected to vrible force F() tht is pplied in the direction of motion. Then we define the work W performed b the force on the object to be W = b F()d () Hooke s lw [Robert Hooke (635 73), English phsicist] sttes tht under pproprite conditions spring tht is stretched units beond its nturl length pulls bck with force F() = k where k is constnt (clled the spring constnt or spring stiffness). The vlue of k depends on such fctors s the thickness of the spring nd the mteril used in its composition. Since k = F()/, the constnt k hs units of force per unit length. Emple 3 length. A spring eerts force of 5 N when stretched mbeond its nturl Nturl position of spring Figure () Find the spring constnt k. (b) How much work is required to stretch the spring.8 m beond its nturl length? Solution (). From Hooke s lw, F() = k From the dt, F() = 5 N when = m,so5= k. Thus, the spring constnt is k = 5 newtons per meter (N/m). This mens tht the force F()required to stretch the spring meters is F() = 5 (3) Solution (b). Plce the spring long coordinte line s shown in Figure We wnt to find the work W required to stretch the spring over the intervl from = to =.8. From () nd (3) the work W required is b.8 ].8 W = F()d = 5d= 5 = 8. J NASA Mrshll Spce Flight Center Collection The Interntionl Spce Sttion hs hd continuous humn occuption since November, nd is coopertive effort of the United Sttes, Russi, Europe, Jpn, nd Cnd. Emple 4 An stronut s weight (or more precisel, Erth weight) is the force eerted on the stronut b the Erth s grvit. As the stronut moves upwrd into spce, the grvittionl pull of the Erth decreses, nd hence so does his or her weight. If the Erth is ssumed to be sphere of rdius 4 mi, then it follows from Newton s Lw of Universl Grvittion tht n stronut who weighs 5 lb on Erth will hve weight of w() =,4,, lb, 4 t distnce of miles from the Erth s center (Eercise 5). Use this formul to estimte the work in foot-pounds required to lift the stronut miles upwrd to the Interntionl Spce Sttion. Solution. Since the Erth hs rdius of 4 mi, the stronut is lifted from point tht is 4 mi from the Erth s center to point tht is 4 mi from the Erth s center. Thus,

41 6.6 Work 453 from (), the work W required to lift the stronut is W = 4 4,4,, d =,4,, ] ,7 + 6, = 3,8 mile-pounds = (3,8 mi lb) (58 ft/mi).65 8 ft lb CALCULATING WORK FROM BASIC PRINCIPLES Some problems cnnot be solved b mechnicll substituting into formuls, nd one must return to bsic principles to obtin solutions. This is illustrted in the net emple. Emple 5 Figure shows conicl continer of rdius ft nd height 3 ft. Suppose tht this continer is filled with wter to depth of 5 ft. How much work is required to pump ll of the wter out through hole in the top of the continer? Solution. Our strteg will be to divide the wter into thin lers, pproimte the work required to move ech ler to the top of the continer, dd the pproimtions for the lers to obtin Riemnn sum tht pproimtes the totl work, nd then tke the limit of the Riemnn sums to produce n integrl for the totl work. To implement this ide, introduce n -is s shown in Figure 6.6.4, nd divide the wter into n lers with k denoting the thickness of the kth ler. This division induces prtition of the intervl [5, 3] into n subintervls. Although the upper nd lower surfces of the kth ler re t different distnces from the top, the difference will be smll if the ler is thin, nd we cn resonbl ssume tht the entire ler is concentrted t single point k (Figure 6.6.4). Thus, the work W k required to move the kth ler to the top of the continer is pproimtel W k F k k (4) where F k is the force required to lift the kth ler. But the force required to lift the kth ler is the force needed to overcome grvit, nd this is the sme s the weight of the ler. If the ler is ver thin, we cn pproimte the volume of the kth ler with the volume of clinder of height k nd rdius r k, where (b similr tringles) r k = 3 = 3 k or, equivlentl, r k = k / 3 (Figure 6.6.4b). Therefore, the volume of the kth ler of wter is pproimtel πr k k = π( k / 3) k = π 9 ( k ) k Since the weight densit of wter is 6.4 lb/ft 3, it follows tht F k 6.4π (k 9 ) k Thus, from (4) ( ) 6.4π W k (k 9 ) k k = 6.4π (k 9 )3 k

42 454 Chpter 6 / Applictions of the Definite Integrl in Geometr, Science, nd Engineering nd hence the work W required to move ll n lers hs the pproimtion n n 6.4π W = W k (k 9 )3 k k= To find the ect vlue of the work we tke the limit s m k. This ields n 6.4π 3 W = lim ( 6.4π k m k 9 )3 k = 3 d 9 = 6.4π 9 ( 4 k= )] k= 5 =,36,5π 4,35, ft lb 5 k * 3 k * 5 ft Δ k 3 k * r k ft 3 Figure () (b) Mike Brinson/Gett Imges The work performed b the skter's stick in brief intervl of time produces the blinding speed of the hocke puck. THE WORK ENERGY RELATIONSHIP When ou see n object in motion, ou cn be certin tht somehow work hs been epended to crete tht motion. For emple, when ou drop stone from building, the stone gthers speed becuse the force of the Erth s grvit is performing work on it, nd when hocke pler strikes puck with hocke stick, the work performed on the puck during the brief period of contct with the stick cretes the enormous speed of the puck cross the ice. However, eperience shows tht the speed obtined b n object depends not onl on the mount of work done, but lso on the mss of the object. For emple, the work required to throw5ozbsebll 5 mi/h would ccelerte lb bowling bll to less thn 9 mi/h. Using the method of substitution for definite integrls, we will derive simple eqution tht reltes the work done on n object to the object s mss nd velocit. Furthermore, this eqution will llow us to motivte n pproprite definition for the energ of motion of n object. As in Definition 6.6.3, we will ssume tht n object moves in the positive direction long coordinte line over the intervl [,b] while subjected to force F() tht is pplied in the direction of motion. We let m denote the mss of the object, nd we let = (t), v = v(t) = (t), nd = (t) = v (t) denote the respective position, velocit, nd ccelertion of the object t time t. We will need the following importnt result from phsics tht reltes the force cting on n object with the mss nd ccelertion of the object newton s second lw of motion If n object with mss m is subjected to force F, then the object undergoes n ccelertion tht stisfies the eqution F = m (5) It follows from Newton s Second Lw of Motion tht F((t)) = m(t) = mv (t)

43 Assume tht with (t ) = nd (t ) = b v(t ) = v i nd v(t ) = v f the initil nd finl velocities of the object, respectivel. Then W = = = = = b t F()d = (t ) (t ) F()d t F ((t)) (t) dt B Theorem 5.9. with = (t),d = (t) dt t t mv (t)v(t) dt = v(t ) v(t ) vf v i mv dv We see from the eqution t mv dv = mv v f v i t mv(t)v (t) dt B Theorem 5.9. with v = v(t), dv = v (t) dt = mv f mv i 6.6 Work 455 W = mv f mv i (6) tht the work done on the object is equl to the chnge in the quntit mv from its initil vlue to its finl vlue. We will refer to Eqution (6) s the work energ reltionship. If we define the energ of motion or kinetic energ of our object to be given b K = mv (7) then Eqution (6) tells us tht the work done on n object is equl to the chnge in the object s kinetic energ. Loosel speking, we m think of work done on n object s being trnsformed into kinetic energ of the object. The units of kinetic energ re the sme s the units of work. For emple, in SI kinetic energ is mesured in joules (J). Emple 6 A spce probe of mss m = 5. 4 kg trvels in deep spce subjected onl to the force of its own engine. Strting t time when the speed of the probe is v =. 4 m/s, the engine is fired continuousl over distnce of.5 6 m with constnt force of 4. 5 N in the direction of motion. Wht is the finl speed of the probe? Solution. Since the force pplied b the engine is constnt nd in the direction of motion, the work W epended b the engine on the probe is W = force distnce = (4. 5 N) (.5 6 m) =. J From (6), the finl kinetic energ K f = mv f of the probe cn be epressed in terms of the work W nd the initil kinetic energ K i = mv i s K f = W + K i Thus, from the known mss nd initil speed we hve K f = (. J) + (5. 4 kg)(. 4 m/s) = 4.5 J The finl kinetic energ is K f = mv f, so the finl speed of the probe is Kf v f = m = (4.5 ).7 4 m/s 5. 4

44 456 Chpter 6 / Applictions of the Definite Integrl in Geometr, Science, nd Engineering QUICK CHECK EXERCISES 6.6 (See pge 458 for nswers.). If constnt force of 5 lb moves n object ft, then the work done b the force on the object is.. A newton-meter is lso clled. A dnecentimeter is lso clled n. 3. Suppose tht n object moves in the positive direction long coordinte line over the intervl [,b]. The work performed on the object b vrible force F()pplied in the direction of motion is W =. 4. A force F() = N pplied in the positive -direction moves n object 3 m from = to = 5. The work done b the force on the object is. EXERCISE SET 6.6 FOCUS ON CONCEPTS. A vrible force F() in the positive -direction is grphed in the ccompning figure. Find the work done b the force on prticle tht moves from = to = 3. Force F (lb) Position (ft) Figure E-. A vrible force F() in the positive -direction is grphed in the ccompning figure. Find the work done b the force on prticle tht moves from = to = 5. Force F (N) Position (m) Figure E- 3. For the vrible force F() in Eercise, consider the distnce d for which the work done b the force on the prticle when the prticle moves from = to = d is hlf of the work done when the prticle moves from = to = 5. B inspecting the grph of F,isd more or less thn.5? Eplin, nd then find the ect vlue of d. 4. Suppose tht vrible force F()is pplied in the positive -direction so tht n object moves from = to = b. Relte the work done b the force on the object nd the verge vlue of F over [,b], nd illustrte this reltionship grphicll. 5. A constnt force of lb in the positive -direction is pplied to prticle whose velocit versus time curve is shown in the ccompning figure. Find the work done b the force on the prticle from time t = tot = 5. Velocit v (ft/s) Time t (s) Figure E-5 6. A spring eerts force of 6 N when it is stretched from its nturl length of 4mtolength of 4 m. Find the work required to stretch the spring from its nturl length to length of 6 m. 7. A spring eerts force of N when it is stretched. m beond its nturl length. How much work is required to stretch the spring.8 m beond its nturl length? 8. A spring whose nturl length is 5 cm eerts force of 45 N when stretched to length of cm. () Find the spring constnt (in newtons/meter). (b) Find the work tht is done in stretching the spring 3 cm beond its nturl length. (c) Find the work done in stretching the spring from length of cm to length of 5 cm. 9. Assume tht ft lb of work is required to stretch spring ft beond its nturl length. Wht is the spring constnt? 3 True Flse Determine whether the sttement is true or flse. Eplin our nswer.. In order to support the weight of prked utomobile, the surfce of drivew must do work ginst the force of grvit on the vehicle.. A force of lb in the direction of motion of n object tht moves 5 ft in s does si times the work of force of lb in the direction of motion of n object tht moves 5 ft in s.. It follows from Hooke s lw tht in order to double the distnce spring is stretched beond its nturl length, four times s much work is required. 3. In the Interntionl Sstem of Units, work nd kinetic energ hve the sme units.

45 6.6 Work A clindricl tnk of rdius 5 ft nd height 9 ft is two-thirds filled with wter. Find the work required to pump ll the wter over the upper rim. 5. Solve Eercise 4 ssuming tht the tnk is hlf-filled with wter. 6. A cone-shped wter reservoir is ft in dimeter cross the top nd 5 ft deep. If the reservoir is filled to depth of ft, how much work is required to pump ll the wter to the top of the reservoir? 7. The vt shown in the ccompning figure contins wter to depth of m. Find the work required to pump ll the wter to the top of the vt. [Use 98 N/m 3 s the weight densit of wter.] 8. The clindricl tnk shown in the ccompning figure is filled with liquid weighing 5 lb/ft 3. Find the work required to pump ll the liquid to level ft bove the top of the tnk. 3 m 6 m Figure E-7 4 m 4 ft ft Figure E-8 9. A swimming pool is built in the shpe of rectngulr prllelepiped ft deep, 5 ft wide, nd ft long. () If the pool is filled to ft below the top, how much work is required to pump ll the wter into drin t the top edge of the pool? (b) A one-horsepower motor cn do 55 ft lb of work per second. Wht size motor is required to empt the pool in hour?. How much work is required to fill the swimming pool in Eercise 9 to ft below the top if the wter is pumped in through n opening locted t the bottom of the pool?. A ft length of steel chin weighing 5 lb/ft is dngling from pulle. How much work is required to wind the chin onto the pulle?. A 3 lb bucket contining lb of wter is hnging t the end of ft rope tht weighs 4 oz/ft. The other end of the rope is ttched to pulle. How much work is required to wind the length of rope onto the pulle, ssuming tht the rope is wound onto the pulle t rte of ft/s nd tht s the bucket is being lifted, wter leks from the bucket t rte of.5 lb/s? 3. A rocket weighing 3 tons is filled with 4 tons of liquid fuel. In the initil prt of the flight, fuel is burned off t constnt rte of tons per ft of verticl height. How much work in foot-tons (ft ton) is done lifting the rocket 3 ft? 4. It follows from Coulomb s lw in phsics tht two like electrosttic chrges repel ech other with force inversel proportionl to the squre of the distnce between them. Suppose tht two chrges A nd B repel with force of k newtons when the re positioned t points A(, ) nd B(,), where is mesured in meters. Find the work W required to move chrge A long the -is to the origin if chrge B remins sttionr. 5. It follows from Newton s Lw of Universl Grvittion tht the grvittionl force eerted b the Erth on n object bove the Erth s surfce vries inversel s the squre of its distnce from the Erth s center. Thus, n object s weight w() is relted to its distnce from the Erth s center b formul of the form w() = k where k is constnt of proportionlit tht depends on the mss of the object. () Use this fct nd the ssumption tht the Erth is sphere of rdius 4 mi to obtin the formul for w() in Emple 4. (b) Find formul for the weight w() of stellite tht is mi from the Erth s surfce if its weight on Erth is 6 lb. (c) How much work is required to lift the stellite from the surfce of the Erth to n orbitl position tht is mi high? 6. () The formul w() = k/ in Eercise 5 is pplicble to ll celestil bodies. Assuming tht the Moon is sphere of rdius 8 mi, find the force tht the Moon eerts on n stronut who is mi from the surfce of the Moon if her weight on the Moon s surfce is lb. (b) How much work is required to lift the stronut to point tht is.8 mi bove the Moon s surfce? 7. The world s first commercil high-speed mgnetic levittion (MAGLEV) trin, 3 km double-trck project connecting Shnghi, Chin, to Pudong Interntionl Airport, begn full revenue service in 3. Suppose tht MAGLEV trin hs mss m = 4. 5 kg nd tht strting t time when the trin hs speed of m/s the engine pplies force of N in the direction of motion over distnce of 3. 3 m. Use the work energ reltionship (6) to find the finl speed of the trin. 8. Assume tht Mrs probe of mss m =. 3 kg is subjected onl to the force of its own engine. Strting t time when the speed of the probe is v =. 4 m/s, the engine is fired continuousl over distnce of.5 5 m with constnt force of. 5 N in the direction of motion. Use the work energ reltionship (6) to find the finl speed of the probe. 9. On August, 97 meteorite with n estimted mss of 4 6 kg nd n estimted speed of 5 km/s skipped cross the tmosphere bove the western United Sttes nd Cnd but fortuntel did not hit the Erth. (cont.)

46 458 Chpter 6 / Applictions of the Definite Integrl in Geometr, Science, nd Engineering () Assuming tht the meteorite hd hit the Erth with speed of 5 km/s, wht would hve been its chnge in kinetic energ in joules (J)? (b) Epress the energ s multiple of the eplosive energ of megton of TNT, which is 4. 5 J. (c) The energ ssocited with the Hiroshim tomic bomb ws 3 kilotons of TNT. To how mn such bombs would the meteorite impct hve been equivlent? 3. Writing After reding Emples 3 5, student clssifies work problems s either pushing/pulling or pumping. Describe these ctegories in our own words nd discuss the methods used to solve ech tpe. Give emples to illustrte tht these ctegories re not mutull eclusive. 3. Writing How might ou recognize tht problem cn be solved b mens of the work energ reltionship? Tht is, wht sort of givens nd unknowns would suggest such solution? Discuss two or three emples. QUICK CHECK ANSWERS ft lb. joule; erg 3. b F()d 4. 9J 6.7 MOMENTS, CENTERS OF GRAVITY, AND CENTROIDS Suppose tht rigid phsicl bod is cted on b constnt grvittionl field. Becuse the bod is composed of mn prticles, ech of which is ffected b grvit, the ction of the grvittionl field on the bod consists of lrge number of forces distributed over the entire bod. However, it is fct of phsics tht these individul forces cn be replced b single force cting t point clled the center of grvit of the bod. In this section we will show how integrls cn be used to locte centers of grvit. Figure 6.7. The thickness of lmin is negligible. The units in Eqution () re consistent since mss = (mss/re) re. (, ) = + DENSITY AND MASS OF A LAMINA Let us consider n idelized flt object tht is thin enough to be viewed s two-dimensionl plne region (Figure 6.7.). Such n object is clled lmin. A lmin is clled homogeneous if its composition is uniform throughout nd inhomogeneous otherwise. We will consider homogeneous lmins in this section. Inhomogeneous lmins will be discussed in Chpter 4. The densit of homogeneous lmin is defined to be its mss per unit re. Thus, the densit δ of homogeneous lmin of mss M nd re A is given b δ = M/A. Notice tht the mss M of homogeneous lmin cn be epressed s M = δa () Emple A tringulr lmin with vertices (, ), (, ), nd (, ) hs densit δ = 3. Find its totl mss. Solution. Referring to () nd Figure 6.7., the mss M of the lmin is M = δa = 3 = 3 (unit of mss) (, ) (, ) Figure 6.7. CENTER OF GRAVITY OF A LAMINA Assume tht the ccelertion due to the force of grvit is constnt nd cts downwrd, nd suppose tht lmin occupies region R in horizontl -plne. It cn be shown tht there eists unique point (,ȳ) (which m or m not belong to R) such tht the effect

47 6.7 Moments, Centers of Grvit, nd Centroids 459 of grvit on the lmin is equivlent to tht of single force cting t the point (,ȳ). This point is clled the center of grvit of the lmin, nd if it is in R, then the lmin will blnce horizontll on the point of support plced t (,ȳ). For emple, the center of grvit of homogeneous disk is t the center of the disk, nd the center of grvit of homogeneous rectngulr region is t the center of the rectngle. For n irregulrl shped homogeneous lmin, locting the center of grvit requires clculus problem Let f be positive continuous function on the intervl [,b]. Suppose tht homogeneous lmin with constnt densit δ occupies region R in horizontl -plne bounded b the grphs of = f(), =, =, nd = b. Find the coordintes (,ȳ) of the center of grvit of the lmin. To motivte the solution, consider wht hppens if we tr to blnce the lmin on knife-edge prllel to the -is. Suppose the lmin in Figure is plced on knifeedge long line = c tht does not pss through the center of grvit. Becuse the lmin behves s if its entire mss is concentrted t the center of grvit (,ȳ), the lmin will be rottionll unstble nd the force of grvit will cuse rottion bout = c. Similrl, the lmin will undergo rottion if plced on knife-edge long = d. However, if the knife-edge runs long the line =ȳ through the center of grvit, the lmin will be in perfect blnce. Similrl, the lmin will be in perfect blnce on knife-edge long the line = through the center of grvit. This suggests tht the center of grvit of lmin cn be determined s the intersection of two lines of blnce, one prllel to the -is nd the other prllel to the -is. In order to find these lines of blnce, we will need some preliminr results bout rottions. = = c = d (, ) = f () b Figure Force of grvit cting on the center of grvit of the lmin Children on seesw lern b eperience tht lighter child cn blnce hevier one b sitting frther from the fulcrum or pivot point. This is becuse the tendenc for n object to produce rottion is proportionl not onl to its mss but lso to the distnce between the object nd the fulcrum. To mke this more precise, consider n -is, which we view s weightless bem. If mss m is locted on the is t, then the tendenc for tht mss to produce rottion of the bem bout point on the is is mesured b the following quntit, clled the moment of m bout = : [ ] moment of m = m( ) bout

48 46 Chpter 6 / Applictions of the Definite Integrl in Geometr, Science, nd Engineering m Positive moment bout (clockwise rottion) m Negtive moment bout (counterclockwise rottion) The number is clled the lever rm. Depending on whether the mss is to the right or left of, the lever rm is either the distnce between nd or the negtive of this distnce (Figure 6.7.4). Positive lever rms result in positive moments nd clockwise rottions, nd negtive lever rms result in negtive moments nd counterclockwise rottions. Suppose tht msses m,m,...,m n re locted t,,..., n on coordinte is nd fulcrum is positioned t the point (Figure 6.7.5). Depending on whether the sum of the moments bout, n m k ( k ) = m ( ) + m ( ) + +m n ( n ) k= is positive, negtive, or zero, weightless bem long the is will rotte clockwise bout, rotte counterclockwise bout, or blnce perfectl. In the lst cse, the sstem of msses is sid to be in equilibrium. Figure Figure Fulcrum m n n m m... c m c = c Figure = R R R 3 3 (* k, * k ) = * k, () f(* k )... (, ) = = f() R n n b = n = f() The preceding ides cn be etended to msses distributed in two-dimensionl spce. If we imgine the -plne to be weightless sheet supporting mss m locted t point (, ), then the tendenc for the mss to produce rottion of the sheet bout the line = is m( ), clled the moment of m bout =, nd the tendenc for the mss to produce rottion bout the line = c is m( c), clled the moment of m bout = c (Figure 6.7.6). In summr, [ moment of m bout the line = ] = m( ) nd [ ] moment of m bout the = m( c) ( 3) line = c If number of msses re distributed throughout the -plne, then the plne (viewed s weightless sheet) will blnce on knife-edge long the line = if the sum of the moments bout the line is zero. Similrl, the plne will blnce on knife-edge long the line = c if the sum of the moments bout tht line re zero. We re now red to solve Problem The bsic ide for solving this problem is to divide the lmin into strips whose res m be pproimted b the res of rectngles. These re pproimtions, long with Formuls () nd (3), will llow us to crete Riemnn sum tht pproimtes the moment of the lmin bout horizontl or verticl line. B tking the limit of Riemnn sums we will then obtin n integrl for the moment of lmin bout horizontl or verticl line. We observe tht since the lmin blnces on the lines = nd =ȳ, the moment of the lmin bout those lines should be zero. This observtion will enble us to clculte nd ȳ. To implement this ide, we divide the intervl [,b] into n subintervls b inserting the points,,..., n between = nd b = n. This hs the effect of dividing the lmin R into n strips R,R,...,R n (Figure 6.7.7). Suppose tht the kth strip etends from k to k nd tht the width of this strip is k = k k Figure (b) b We will let k be the midpoint of the kth subintervl nd we will pproimte R k b rectngle of width k nd height f(k ). From (), the mss M k of this rectngle is M k = δf (k ) k, nd we will ssume tht the rectngle behves s if its entire mss is concentrted t its center (k, k ) = ( k, f( k )) (Figure 6.7.7b). It then follows from () nd (3) tht the moments of R k bout the lines = nd =ȳ m be pproimted

49 6.7 Moments, Centers of Grvit, nd Centroids 46 b (k ) M k nd (k ȳ) M k, respectivel. Adding these pproimtions ields the following Riemnn sums tht pproimte the moment of the entire lmin bout the lines = nd =ȳ: n n (k ) M k = (k )δf( k ) k k= n (k ȳ) M k = k= k= n ( f( k ) k= ) ȳ δf (k ) k Tking the limits s n increses nd the widths of ll the rectngles pproch zero ields the definite integrls b b ( ) f() ( )δf() d nd ȳ δf() d tht represent the moments of the lmin bout the lines = nd =ȳ. Since the lmin blnces on those lines, the moments of the lmin bout those lines should be zero: b b ( ) f() ( )δf() d = ȳ δf() d = Since nd ȳ re constnt, these equtions cn be rewritten s b b δf() d = δ(f()) d =ȳ b b δf() d δf() d from which we obtin the following formuls for the center of grvit of the lmin: = Center of Grvit (, ȳ) of Lmin b b δf() d, ȳ = δf() d b δ (f()) d b δf() d (4 5) Observe tht in both formuls the denomintor is the mss M of the lmin. The numertor in the formul for is denoted b M nd is clled the first moment of the lmin bout the -is; the numertor of the formul for ȳ is denoted b M nd is clled the first moment of the lmin bout the -is. Thus, we cn write (4) nd (5) s Alterntive Formuls for Center of Grvit (, ȳ) of Lmin = M M = b δf() d mss of R ȳ = M M = mss of R b (6) δ (f()) d (7) Emple Find the center of grvit of the tringulr lmin with vertices (, ), (, ), nd (, ) nd densit δ = 3. Solution. The lmin is shown in Figure In Emple we found the mss of the lmin to be M = 3

50 46 Chpter 6 / Applictions of the Definite Integrl in Geometr, Science, nd Engineering The moment of the lmin bout the -is is M = = δf() d = ( 3 + 3)d = nd the moment bout the -is is M = = From (6) nd (7), δ(f()) d = 3 ( + )d = 3 so the center of grvit is ( 3, 3). 3( + )d ( )] = + 3 = 3 ( + ) d ( )] = 3 = M M = / 3/ = 3, ȳ = M M = / 3/ = 3 ( ) = 3 In the cse of homogeneous lmin, the center of grvit of lmin occuping the region R is clled the centroid of the region R. Since the lmin is homogeneous, δ is constnt. The fctor δ in (4) nd (5) m thus be moved through the integrl signs nd cnceled, nd (4) nd (5) cn be epressed s Since the densit fctor hs cnceled, we m interpret the centroid s geometric propert of the region, nd distinguish it from the center of grvit, which is phsicl propert of n idelized object tht occupies the region. = ȳ = b b b b Centroid of Region R f() d f()d = (f()) d f()d re of R = b f() d (8) b re of R (f()) d (9) Emple 3 Find the centroid of the semicirculr region in Figure Figure R Solution. B smmetr, = since the -is is obviousl line of blnce. To find ȳ, first note tht the eqution of the semicircle is = f() =. From (9), ȳ = re of R (f()) d = π ( )d = π = π = π ( 3 3 )] [( 3 3 ) 3 ( )] 3 ( 4 3 ) 3 = 4 3π so the centroid is (, 4/3π).

51 6.7 Moments, Centers of Grvit, nd Centroids 463 (* k, * k ) = * k, ( f (* k ) + g(* k )) = f () Figure = g() OTHER TYPES OF REGIONS The strteg used to find the center of grvit of the region in Problem 6.7. cn be used to find the center of grvit of regions tht re not of tht form. Consider homogeneous lmin tht occupies the region R between two continuous functions f()nd g() over the intervl [,b], where f() g() for b. To find the center of grvit of this lmin we cn subdivide it into n strips using lines prllel to the -is. If k is the midpoint of the kth strip, the strip cn be pproimted b rectngle of width k nd height f(k ) g( k ). We ssume tht the entire mss of the kth rectngle is concentrted t its center (k, k ) = ( k, (f( k ) + g( k ))) (Figure 6.7.9). Continuing the rgument s in the solution of Problem 6.7., we find tht the center of grvit of the lmin is = ȳ = b b b (f() g()) d (f() g()) d = re of R ( [f()] [g()] ) d b = (f() g()) d b (f() g()) d () b ( [f()] [g()] ) d () re of R Note tht the densit of the lmin does not pper in Equtions () nd (). This reflects the fct tht the centroid is geometric propert of R. Emple 4 = + 6. Find the centroid of the region R enclosed between the curves = nd = (3, 9) (, 4) 4 = Figure 6.7. Solution. To begin, we note tht the two curves intersect when = nd = 3 nd tht + 6 over tht intervl (Figure 6.7.). The re of R is From () nd (), nd 3 = re of R = 6 5 [( + 6) ] d = [( + 6) ] d ( = = ȳ = re of R = = 6 5 so the centroid of R is (, 4). 3 = = 4 )] 3 (( + 6) ( ) )d ( )d ( )] 3

52 464 Chpter 6 / Applictions of the Definite Integrl in Geometr, Science, nd Engineering Figure 6.7. (* k, * k ) = w(* k ), * k = w() Suppose tht w is continuous function of on n intervl [c, d] with w() for c d. Consider lmin tht occupies region R bounded bove b = d, below b = c, on the left b the -is, nd on the right b = w() (Figure 6.7.). To find the center of grvit of this lmin, we note tht the roles of nd in Problem 6.7. hve been reversed. We now imgine the lmin to be subdivided into n strips using lines prllel to the -is. We let k be the midpoint of the kth subintervl nd pproimte the strip b rectngle of width k nd height w(k ). We ssume tht the entire mss of the kth rectngle is concentrted t its center (k, k ) = ( w( k ), k ) (Figure 6.7.). Continuing the rgument s in the solution of Problem 6.7., we find tht the center of grvit of the lmin is = ȳ = d c d c d c d c (w()) d w() d w() d w() d = = d re of R c (w()) d () d w() d (3) re of R c Once gin, the bsence of the densit in Equtions () nd (3) reflects the geometric nture of the centroid. R = 3 4 Figure 6.7. Emple 5 Find the centroid of the region R enclosed between the curves =, =, =, nd the -is (Figure 6.7.). Solution. From () nd (3), Note tht = w() = nd tht the re of R is = ȳ = re of R re of R d = 7 3 ( ) d = 3 7 ( )d = ] 5 ] so the centroid of R is (93/7, 45/8) (.39,.67). = = 93 7 = = 45 8 THEOREM OF PAPPUS The following theorem, due to the Greek mthemticin Pppus, gives n importnt reltionship between the centroid of plne region R nd the volume of the solid generted when the region is revolved bout line theorem (Theorem of Pppus) If R is bounded plne region nd L is line tht lies in the plne of R such tht R is entirel on one side of L, then the volume of the solid formed b revolving R bout L is given b volume = (re of R) ( ) distnce trveled b the centroid

53 6.7 Moments, Centers of Grvit, nd Centroids 465 proof We prove this theorem in the specil cse where L is the -is, the region R is in the first qudrnt, nd the region R is of the form given in Problem (A more generl proof will be outlined in the Eercises of Section 4.8.) In this cse, the volume V of the solid formed b revolving R bout L cn be found b the method of clindricl shells (Section 6.3) to be b V = π f() d Thus, it follows from (8) tht V = π [re of R] This completes the proof since π is the distnce trveled b the centroid when R is revolved bout the -is. b Emple 6 Use Pppus Theorem to find the volume V of the torus generted b revolving circulr region of rdius b bout line t distnce (greter thn b) from the center of the circle (Figure 6.7.3). Figure The centroid trvels distnce c. Solution. B smmetr, the centroid of circulr region is its center. Thus, the distnce trveled b the centroid is π. Since the re of circle of rdius b is πb, it follows from Pppus Theorem tht the volume of the torus is V = (π)(πb ) = π b QUICK CHECK EXERCISES 6.7 (See pge 467 for nswers.). The totl mss of homogeneous lmin of re A nd densit δ is.. A homogeneous lmin of mss M nd densit δ occupies region in the -plne bounded b the grphs of = f(), =, =, nd = b, where f is nonnegtive continuous function defined on n intervl [,b]. The -coordinte of the center of grvit of the lmin is M /M, where M is clled the nd is given b the integrl. 3. Let R be the region between the grphs of = nd = for. The re of R is 7 nd the centroid of R is If the region R in Quick Check Eercise 3 is used to generte solid G b rotting R bout horizontl line 6 units bove its centroid, then the volume of G is. EXERCISE SET 6.7 C CAS FOCUS ON CONCEPTS. Msses m = 5, m =, nd m 3 = re positioned on weightless bem s shown in the ccompning figure. () Suppose tht the fulcrum is positioned t = 5. Without computing the sum of moments bout 5, determine whether the sum is positive, zero, or negtive. Eplin. (b) Where should the fulcrum be plced so tht the bem is in equilibrium? m 5 Figure E- m m 3 5 Pppus of Alendri (4th centur A.D.) Greek mthemticin. Pppus lived during the erl Christin er when mthemticl ctivit ws in period of decline. His min contributions to mthemtics ppered in series of eight books clled The Collection (written bout 34 A.D.). This work, which survives onl prtill, contined some originl results but ws devoted mostl to sttements, refinements, nd proofs of results b erlier mthemticins. Pppus Theorem, stted without proof in Book VII of The Collection, ws probbl known nd proved in erlier times. This result is sometimes clled Guldin s Theorem in recognition of the Swiss mthemticin, Pul Guldin ( ), who rediscovered it independentl.

54 466 Chpter 6 / Applictions of the Definite Integrl in Geometr, Science, nd Engineering. Msses m =, m = 3, m 3 = 4, nd m re positioned on weightless bem, with the fulcrum positioned t point 4, s shown in the ccompning figure. () Suppose tht m = 4. Without computing the sum of the moments bout 4, determine whether the sum is positive, zero, or negtive. Eplin. (b) For wht vlue of m is the bem in equilibrium? m m m 3 3 4? 3 4 Figure E- 3 6 Find the centroid of the region b inspection nd confirm our nswer b integrting (, ) m 6. The tringle with vertices (, ), (, ), nd (, ). 3. The region bounded b the grphs of = nd + = The region bounded on the left b the -is, on the right b the line =, below b the prbol =, nd bove b the line = The region bounded b the grphs of = nd = The region bounded b the grphs of = nd =. 7. The region bounded b the grphs of = nd =. 8. The region bounded b the grphs of = /, =, =, nd =. 9. The region bounded b the grphs of =, = /, nd =.. The region bounded b the grphs of = 4 nd + = 5. FOCUS ON CONCEPTS. Use smmetr considertions to rgue tht the centroid of n isosceles tringle lies on the medin to the bse of the tringle.. Use smmetr considertions to rgue tht the centroid of n ellipse lies t the intersection of the mjor nd minor es of the ellipse Find the mss nd center of grvit of the lmin with densit δ. (, ) 3. A lmin bounded b the -is, the line =, nd the curve = ; δ =. 4. A lmin bounded b the grph of = 4 nd the line = ; δ = A lmin bounded b the grph of = nd the line = ; δ = 3. 7 Find the centroid of the region A lmin bounded b the -is nd the grph of the eqution = ; δ = = =. = = C 7 3 Use CAS to find the mss nd center of grvit of the lmin with densit δ. 7. A lmin bounded b = sin, =, =, nd = π; δ = A lmin bounded b = e, =, =, nd = ; δ = /(e ). 9. A lmin bounded b the grph of = ln, the -is, nd the line = ; δ =. 3. A lmin bounded b the grphs of = cos, = sin, =, nd = π/4; δ = +. =. The tringle with vertices (, ), (, ), nd (, ) True Flse Determine whether the sttement is true or flse. Eplin our nswer. [In Eercise 34, ssume tht the (rotted) squre lies in the -plne to the right of the -is.]

55 6.8 Fluid Pressure nd Force The centroid of rectngle is the intersection of the digonls of the rectngle. 3. The centroid of rhombus is the intersection of the digonls of the rhombus. 33. The centroid of n equilterl tringle is the intersection of the medins of the tringle. 34. B rotting squre bout its center, it is possible to chnge the volume of the solid of revolution generted b revolving the squre bout the -is. 35. Find the centroid of the tringle with vertices (, ), (, b), nd (, b). 36. Prove tht the centroid of tringle is the point of intersection of the three medins of the tringle. [Hint: Choose coordintes so tht the vertices of the tringle re locted t (, ), (,), nd (b, c).] 37. Find the centroid of the isosceles trpezoid with vertices (,), (, ), ( b, c), nd (b, c). 38. Prove tht the centroid of prllelogrm is the point of intersection of the digonls of the prllelogrm. [Hint: Choose coordintes so tht the vertices of the prllelogrm re locted t (, ), (,), (b, c), nd (b, + c).] 39. Use the Theorem of Pppus nd the fct tht the volume of sphere of rdius is V = 4 3 π3 to show tht the centroid of the lmin tht is bounded b the -is nd the semicircle = is (, 4/(3π)). (This problem ws solved directl in Emple 3.) 4. Use the Theorem of Pppus nd the result of Eercise 39 to find the volume of the solid generted when the region bounded b the -is nd the semicircle = is revolved bout () the line = (b) the line =. 4. Use the Theorem of Pppus nd the fct tht the re of n ellipse with semies nd b is πb to find the volume of the ellipticl torus generted b revolving the ellipse ( k) + b = bout the -is. Assume tht k>. 4. Use the Theorem of Pppus to find the volume of the solid tht is generted when the region enclosed b = nd = 8 is revolved bout the -is. 43. Use the Theorem of Pppus to find the centroid of the tringulr region with vertices (, ), (, ), nd (,b), where >nd b>. [Hint: Revolve the region bout the - is to obtin ȳ nd bout the -is to obtin.] 44. Writing Suppose tht region R in the plne is decomposed into two regions R nd R whose res re A nd A, respectivel, nd whose centroids re (, ȳ ) nd (, ȳ ), respectivel. Investigte the problem of epressing the centroid of R in terms of A, A, (, ȳ ), nd (, ȳ ). Write short report on our investigtions, supporting our resoning with plusible rguments. Cn ou etend our results to decompositions of R into more thn two regions? 45. Writing How might ou recognize tht problem cn be solved b mens of the Theorem of Pppus? Tht is, wht sort of givens nd unknowns would suggest such solution? Discuss two or three emples. QUICK CHECK ANSWERS 6.7. δa. first moment bout the -is; b δf() d 3. ( 5 4, 3 ) π 6.8 FLUID PRESSURE AND FORCE In this section we will use the integrtion tools developed in the preceding chpter to stud the pressures nd forces eerted b fluids on submerged objects. WHATISAFLUID? A fluid is substnce tht flows to conform to the boundries of n continer in which it is plced. Fluids include liquids, such s wter, oil, nd mercur, s well s gses, such s helium, ogen, nd ir. The stud of fluids flls into two ctegories: fluid sttics (the stud of fluids t rest) nd fluid dnmics (the stud of fluids in motion). In this section we will be concerned onl with fluid sttics; towrd the end of this tet we will investigte problems in fluid dnmics.

56 468 Chpter 6 / Applictions of the Definite Integrl in Geometr, Science, nd Engineering THE CONCEPT OF PRESSURE The effect tht force hs on n object depends on how tht force is spred over the surfce of the object. For emple, when ou wlk on soft snow with boots, the weight of our bod crushes the snow nd ou sink into it. However, if ou put on pir of snowshoes to spred the weight of our bod over greter surfce re, then the weight of our bod hs less of crushing effect on the snow. The concept tht ccounts for both the mgnitude of force nd the re over which it is pplied is clled pressure definition If force of mgnitude F is pplied to surfce of re A, then we define the pressure P eerted b the force on the surfce to be Brnd X Pictures/Gett Imges, Inc. Snowshoes prevent the womn from sinking b spreding her weight over lrge re to reduce her pressure on the snow. P = F A It follows from this definition tht pressure hs units of force per unit re. The most common units of pressure re newtons per squre meter (N/m ) in SI nd pounds per squre inch (lb/in ) or pounds per squre foot (lb/ft ) in the BE sstem. As indicted in Tble 6.8., one newton per squre meter is clled pscl (P). A pressure of P is quite smll (P=.45 4 lb/in ), so in countries using SI, tire pressure guges re usull clibrted in kilopscls (kp), which is pscls. () sstem Tble 6.8. units of force nd pressure force re = pressure SI BE BE newton (N) pound (lb) pound (lb) squre meter (m ) squre foot (ft ) squre inch (in ) pscl (P) lb/ft lb/in (psi) conversion fctors: P.45 4 lb/in.9 lb/ft lb/in P lb/ft 47.9 P Blise Pscl (63 66) French mthemticin nd scientist. Pscl s mother died when he ws three ers old nd his fther, highl educted mgistrte, personll provided the bo s erl eduction. Although Pscl showed n inclintion for science nd mthemtics, his fther refused to tutor him in those subjects until he mstered Ltin nd Greek. Pscl s sister nd primr biogrpher climed tht he independentl discovered the first thirt-two propositions of Euclid without ever reding book on geometr. (However, it is generll greed tht the stor is pocrphl.) Nevertheless, the precocious Pscl published highl respected ess on conic sections b the time he ws siteen ers old. Descrtes, who red the ess, thought it so brillint tht he could not believe tht it ws written b such oung mn. B ge 8 his helth begn to fil nd until his deth he ws in frequent pin. However, his cretivit ws unimpired. Pscl s contributions to phsics include the discover tht ir pressure decreses with ltitude nd the principle of fluid pressure tht bers his nme. However, the originlit of his work is questioned b some historins. Pscl mde mjor contributions to brnch of mthemtics clled projective geometr, nd he helped to develop probbilit theor through series of letters with Fermt. In 646, Pscl s helth problems resulted in deep emotionl crisis tht led him to become incresingl concerned with religious mtters. Although born Ctholic, he converted to religious doctrine clled Jnsenism nd spent most of his finl ers writing on religion nd philosoph. [Imge:

57 6.8 Fluid Pressure nd Force 469 In this section we will be interested in pressures nd forces on objects submerged in fluids. Pressures themselves hve no directionl chrcteristics, but the forces tht the crete lws ct perpendiculr to the fce of the submerged object. Thus, in Figure 6.8. the wter pressure cretes horizontl forces on the sides of the tnk, verticl forces on the bottom of the tnk, nd forces tht vr in direction, so s to be perpendiculr to the different prts of the swimmer s bod. Emple Referring to Figure 6.8., suppose tht the bck of the swimmer s hnd hs surfce re of m nd tht the pressure cting on it is 5. 4 P ( relistic vlue ner the bottom of deep diving pool). Find the force tht cts on the swimmer s hnd. Solution. From (), the force F is Fluid forces lws ct perpendiculr to the surfce of submerged object. Figure 6.8. F = PA = (5. 4 N/m )(8.4 3 m ) 4.3 N This is quite lrge force (nerl lb in the BE sstem). si Mchine oil Gsoline Fresh wter Sewter Mercur be sstem Mchine oil Gsoline Fresh wter Sewter Mercur Tble 6.8. weight densities N/m ,45 33,46 lb/ft All densities re ffected b vritions in temperture nd pressure. Weight densities re lso ffected b vritions in g. FLUID DENSITY Scub divers know tht the pressure nd forces on their bodies increse with the depth the dive. This is cused b the weight of the wter nd ir bove the deeper the diver goes, the greter the weight bove nd so the greter the pressure nd force eerted on the diver. To clculte pressures nd forces on submerged objects, we need to know something bout the chrcteristics of the fluids in which the re submerged. For simplicit, we will ssume tht the fluids under considertion re homogeneous, b which we men tht n two smples of the fluid with the sme volume hve the sme mss. It follows from this ssumption tht the mss per unit volume is constnt δ tht depends on the phsicl chrcteristics of the fluid but not on the size or loction of the smple; we cll δ = m V the mss densit of the fluid. Sometimes it is more convenient to work with weight per unit volume thn with mss per unit volume. Thus, we define the weight densit ρ of fluid to be ρ = w V where w is the weight of fluid smple of volume V. Thus, if the weight densit of fluid is known, then the weight w of fluid smple of volume V cn be computed from the formul w = ρv. Tble 6.8. shows some tpicl weight densities. () (3) A h h FLUID PRESSURE To clculte fluid pressures nd forces we will need to mke use of n eperimentl observtion. Suppose tht flt surfce of re A is submerged in homogeneous fluid of weight densit ρ such tht the entire surfce lies between depths h nd h, where h h (Figure 6.8.). Eperiments show tht on both sides of the surfce, the fluid eerts force tht is perpendiculr to the surfce nd whose mgnitude F stisfies the inequlities ρh A F ρh A (4) Figure 6.8. Thus, it follows from () tht the pressure P = F /A on given side of the surfce stisfies the inequlities ρh P ρh (5)

58 47 Chpter 6 / Applictions of the Definite Integrl in Geometr, Science, nd Engineering F m 6 m The fluid force is the fluid pressure times the re. Figure Note tht it is now strightforwrd mtter to clculte fluid force nd pressure on flt surfce tht is submerged horizontll t depth h, for then h = h = h nd inequlities (4) nd (5) become the equlities F = ρha (6) nd P = ρh (7) Emple Find the fluid pressure nd force on the top of flt circulr plte of rdius m tht is submerged horizontll in wter t depth of 6 m (Figure 6.8.3). Solution. Since the weight densit of wter is ρ = 98 N/m 3, it follows from (7) tht the fluid pressure is P = ρh = (98)(6) = 58,86 P nd it follows from (6) tht the fluid force is F = ρha = ρh(πr ) = (98)(6)(4π) = 35,44π 739,7 N FLUID FORCE ON A VERTICAL SURFACE It ws es to clculte the fluid force on the horizontl plte in Emple becuse ech point on the plte ws t the sme depth. The problem of finding the fluid force on verticl surfce is more complicted becuse the depth, nd hence the pressure, is not constnt over the surfce. To find the fluid force on verticl surfce we will need clculus. b w() h() 6.8. problem Suppose tht flt surfce is immersed verticll in fluid of weight densit ρ nd tht the submerged portion of the surfce etends from = to = b long n -is whose positive direction is down (Figure 6.8.4). For b, suppose tht w() is the width of the surfce nd tht h() is the depth of the point. Define wht is ment b the fluid force F on the surfce, nd find formul for computing it. () = 3. n b = n A n (b) A A A 3. The bsic ide for solving this problem is to divide the surfce into horizontl strips whose res m be pproimted b res of rectngles. These re pproimtions, long with inequlities (4), will llow us to crete Riemnn sum tht pproimtes the totl force on the surfce. B tking limit of Riemnn sums we will then obtin n integrl for F. To implement this ide, we divide the intervl [,b] into n subintervls b inserting the points,,..., n between = nd b = n. This hs the effect of dividing the surfce into n strips of re A k,k =,,...,n (Figure 6.8.4b). It follows from (4) tht the force F k on the kth strip stisfies the inequlities k * k k Δ k h(* k ) or, equivlentl, ρh( k )A k F k ρh( k )A k h( k ) F k ρa k h( k ) b Figure w( k *) (c) Since the depth function h() increses linerl, there must eist point k between k nd k such tht h( k ) = F k ρa k or, equivlentl, F k = ρh( k )A k

59 6.8 Fluid Pressure nd Force 47 We now pproimte the re A k of the kth strip of the surfce b the re of rectngle of width w(k ) nd height k = k k (Figure 6.8.4c). It follows tht F k m be pproimted s F k = ρh(k )A k ρh(k ) w( k ) k }{{} Are of rectngle Adding these pproimtions ields the following Riemnn sum tht pproimtes the totl force F on the surfce: n n F = F k ρh(k )w( k ) k k= k= Tking the limit s n increses nd the widths of ll the subintervls pproch zero ields the definite integrl n b F = lim ρh(k m k )w( k ) k = ρh()w() d k= In summr, we hve the following result. ft definition Suppose tht flt surfce is immersed verticll in fluid of weight densit ρ nd tht the submerged portion of the surfce etends from = to = b long n -is whose positive direction is down (Figure 6.8.4). For b, suppose tht w() is the width of the surfce nd tht h() is the depth of the point. Then we define the fluid force F on the surfce to be ft () F = b ρh()w() d (8) h() w() = Emple 3 The fce of dm is verticl rectngle of height ft nd width ft (Figure 6.8.5). Find the totl fluid force eerted on the fce when the wter surfce is level with the top of the dm. (b) Figure ft ft () h() = 3 + w() (b) Figure ft 4 Solution. Introduce n -is with its origin t the wter surfce s shown in Figure 6.8.5b. At point on this is, the width of the dm in feet is w() = nd the depth in feet is h() =. Thus, from (8) with ρ = 6.4 lb/ft 3 (the weight densit of wter) we obtin s the totl force on the fce F = (6.4)()()d =,48 =,48 d ] = 6,4, lb Emple 4 A plte in the form of n isosceles tringle with bse ft nd ltitude 4 ft is submerged verticll in mchine oil s shown in Figure Find the fluid force F ginst the plte surfce if the oil hs weight densit ρ = 3 lb/ft 3. Solution. Introduce n -is s shown in Figure 6.8.6b. B similr tringles, the width of the plte, in feet, t depth of h() = (3 + ) ft stisfies w() = 4, so w() = 5

60 47 Chpter 6 / Applictions of the Definite Integrl in Geometr, Science, nd Engineering Thus, it follows from (8) tht the force on the plte is b 4 ( 5 F = ρh()w() d = (3)(3 + ) = 75 4 [ 3 (3 + )d = ] 4 ) d = 34 lb QUICK CHECK EXERCISES 6.8 (See pge 473 for nswers.). The pressure unit equivlent to newton per squre meter (N/m ) is clled. The pressure unit psi stnds for.. Given tht the weight densit of wter is 98 N/m 3, the fluid pressure on rectngulr m 3 m flt plte submerged horizontll in wter t depth of m is. The fluid force on the plte is. 3. Suppose tht flt surfce is immersed verticll in fluid of weight densit ρ nd tht the submerged portion of the surfce etends from = to = b long n -is whose positive direction is down. If, for b, the surfce hs width w() nd depth h(), then the fluid force on the surfce is F =. 4. A rectngulr plte m wide nd 3 m high is submerged verticll in wter so tht the top of the plte is 5 m below the wter surfce. An integrl epression for the force of the wter on the plte surfce is F =. EXERCISE SET 6.8 In this eercise set, refer to Tble 6.8. for weight densities of fluids, where needed.. A flt rectngulr plte is submerged horizontll in wter. () Find the force (in lb) nd the pressure (in lb/ft )on the top surfce of the plte if its re is ft nd the surfce is t depth of 5 ft. (b) Find the force (in N) nd the pressure (in P) on the top surfce of the plte if its re is 5 m nd the surfce is t depth of m.. () Find the force (in N) on the deck of sunken ship if its re is 6 m nd the pressure cting on it is 6. 5 P. (b) Find the force (in lb) on diver s fce msk if its re is 6 in nd the pressure cting on it is lb/in. 3 8 The flt surfces shown re submerged verticll in wter. Find the fluid force ginst ech surfce. 3. ft 4. m 4 ft m 4 m 5. m 6. 4 ft 4 ft 4 ft 7. 6 m m 8. 8 m m 4 ft 8 ft 6 ft 9. Suppose tht flt surfce is immersed verticll in fluid of weight densit ρ. Ifρ is doubled, is the force on the plte lso doubled? Eplin our resoning.. An oil tnk is shped like right circulr clinder of dimeter 4 ft. Find the totl fluid force ginst one end when the is is horizontl nd the tnk is hlf filled with oil of weight densit 5 lb/ft 3.. A squre plte of side feet is dipped in liquid of weight densit ρ lb/ft 3. Find the fluid force on the plte if verte is t the surfce nd digonl is perpendiculr to the surfce. 5 True Flse Determine whether the sttement is true or flse. Eplin our nswer.. In the Interntionl Sstem of Units, pressure nd force hve the sme units. 3. In clindricl wter tnk (with verticl is), the fluid force on the bse of the tnk is equl to the weight of wter in the tnk. 4. In rectngulr wter tnk, the fluid force on n side of the tnk must be less thn the fluid force on the bse of the tnk. 4 ft

61 6.8 Fluid Pressure nd Force In n wter tnk with flt bse, no mtter wht the shpe of the tnk, the fluid force on the bse is t most equl to the weight of wter in the tnk. 6 9 Formul (8) gives the fluid force on flt surfce immersed verticll in fluid. More generll, if flt surfce is immersed so tht it mkes n ngle of θ<π/ with the verticl, then the fluid force on the surfce is given b F = b ρh()w() sec θd Use this formul in these eercises. 6. Derive the formul given bove for the fluid force on flt surfce immersed t n ngle in fluid. 7. The ccompning figure shows rectngulr swimming pool whose bottom is n inclined plne. Find the fluid force on the bottom when the pool is filled to the top. 4 ft 6 ft 8 ft ft Figure E-7 8. B how mn feet should the wter in the pool of Eercise 7 be lowered in order for the force on the bottom to be reduced b fctor of? 9. The ccompning figure shows dm whose fce is n inclined rectngle. Find the fluid force on the fce when the wter is level with the top of this dm. m m 6 Figure E-9. An observtion window on submrine is squre with ft sides. Using ρ for the weight densit of sewter, find the fluid force on the window when the submrine hs descended so tht the window is verticl nd its top is t depth of h feet. FOCUS ON CONCEPTS. () Show: If the submrine in Eercise descends verticll t constnt rte, then the fluid force on the window increses t constnt rte. (b) At wht rte is the force on the window incresing if the submrine is descending verticll t ft/min?. () Let D = D denote disk of rdius submerged in fluid of weight densit ρ such tht the center of D is h units below the surfce of the fluid. For ech vlue of r in the intervl (,], let D r denote the disk of rdius r tht is concentric with D. Select side of the disk D nd define P(r) to be the fluid pressure on the chosen side of D r. Use (5) to prove tht lim P(r) = ρh r + (b) Eplin wh the result in prt () m be interpreted to men tht fluid pressure t given depth is the sme in ll directions. (This sttement is one version of result known s Pscl s Principle.) 3. Writing Suppose tht we model the Erth s tmosphere s fluid. Atmospheric pressure t se level is P = 4.7 lb/in nd the weight densit of ir t se level is bout ρ = lb/in 3. With these numbers, wht would Formul (7) ield s the height of the tmosphere bove the Erth? Do ou think this nswer is resonble? If not, eplin how we might modif our ssumptions to ield more plusible nswer. 4. Writing Suppose tht the weight densit ρ of fluid is function ρ = ρ() of the depth within the fluid. How do ou think tht Formul (7) for fluid pressure will need to be modified? Support our nswer with plusible rguments. QUICK CHECK ANSWERS 6.8. pscl; pounds per squre inch. 98, P; 588,6 N 3. b ρh()w() d [(5 + )] d

62 474 Chpter 6 / Applictions of the Definite Integrl in Geometr, Science, nd Engineering 6.9 HYPERBOLIC FUNCTIONS AND HANGING CABLES In this section we will stud certin combintions of e nd e, clled hperbolic functions. These functions, which rise in vrious engineering pplictions, hve mn properties in common with the trigonometric functions. This similrit is somewht surprising, since there is little on the surfce to suggest tht there should be n reltionship between eponentil nd trigonometric functions. This is becuse the reltionship occurs within the contet of comple numbers, topic which we will leve for more dvnced courses. DEFINITIONS OF HYPERBOLIC FUNCTIONS To introduce the hperbolic functions, observe from Eercise 65 in Section. tht the function e cn be epressed in the following w s the sum of n even function nd n odd function: e = e + e } {{ } Even + e e }{{} Odd These functions re sufficientl importnt tht there re nmes nd nottion ssocited with them: the odd function is clled the hperbolic sine of nd the even function is clled the hperbolic cosine of. The re denoted b sinh = e e nd cosh = e + e where sinh is pronounced cinch nd cosh rhmes with gosh. From these two building blocks we cn crete four more functions to produce the following set of si hperbolic functions. The terms tnh, sech, nd csch re pronounced tnch, seech, nd coseech, respectivel definition Hperbolic sine Hperbolic cosine Hperbolic tngent sinh = e e cosh = e + e tnh = sinh cosh = e e e + e Hperbolic cotngent coth = cosh sinh = e + e e e Hperbolic secnt sech = cosh = e + e Hperbolic cosecnt csch = sinh = e e TECHNOLOGY MASTERY Computer lgebr sstems hve builtin cpbilities for evluting hperbolic functions directl, but some clcultors do not. However, if ou need to evlute hperbolic function on clcultor, ou cn do so b epressing it in terms of eponentil functions, s in Emple. Emple sinh = e e = = cosh = e + e = + = sinh = e e 3.669

63 6.9 Hperbolic Functions nd Hnging Cbles 475 GRAPHS OF THE HYPERBOLIC FUNCTIONS The grphs of the hperbolic functions, which re shown in Figure 6.9., cn be generted with grphing utilit, but it is worthwhile to observe tht the generl shpe of the grph of = cosh cn be obtined b sketching the grphs of = e nd = e seprtel nd dding the corresponding -coordintes [see prt () of the figure]. Similrl, the generl shpe of the grph of = sinh cn be obtined b sketching the grphs of = e nd = e seprtel nd dding corresponding -coordintes [see prt (b) of the figure]. = e = e = e = e = cosh () = sinh (b) = tnh (c) = coth = sech = csch (d) (e) (f ) Figure 6.9. Glen Allison/Stone/Gett Imges The design of the Gtew Arch in St. Louis is bsed on n inverted hperbolic cosine curve (Eercise 73). Observe tht sinh hs domin of (, + ) nd rnge of (, + ), wheres cosh hs domin of (, + ) nd rnge of [, + ). Observe lso tht = e nd = e re curviliner smptotes for = cosh in the sense tht the grph of = cosh gets closer nd closer to the grph of = e s + nd gets closer nd closer to the grph of = e s. (See Section 4.3.) Similrl, = e is curviliner smptote for = sinh s + nd = e is curviliner smptote s. Other properties of the hperbolic functions re eplored in the eercises. HANGING CABLES AND OTHER APPLICATIONS Hperbolic functions rise in vibrtor motions inside elstic solids nd more generll in mn problems where mechnicl energ is grdull bsorbed b surrounding medium. The lso occur when homogeneous, fleible cble is suspended between two points, s with telephone line hnging between two poles. Such cble forms curve, clled ctenr (from the Ltin cten, mening chin ). If, s in Figure 6.9., coordinte sstem is introduced so tht the low point of the cble lies on the -is, then it cn be shown using principles of phsics tht the cble hs n eqution of the form ( ) = cosh + c

64 476 Chpter 6 / Applictions of the Definite Integrl in Geometr, Science, nd Engineering Figure 6.9. = cosh (/) + c Lrr Auipp/Mir.com/Digitl Rilrod, Inc. A fleible cble suspended between two poles forms ctenr. HYPERBOLIC IDENTITIES The hperbolic functions stisf vrious identities tht re similr to identities for trigono- metric functions. The most fundmentl of these is where the prmeters nd c re determined b the distnce between the poles nd the composition of the cble. which cn be proved b writing cosh sinh = () cosh sinh = (cosh + sinh )(cosh sinh ) ( e + e = + e e )( e + e = e e = e e ) Other hperbolic identities cn be derived in similr mnner or, lterntivel, b performing lgebric opertions on known identities. For emple, if we divide () b cosh, we obtin tnh = sech nd if we divide () b sinh, we obtin coth = csch The following theorem summrizes some of the more useful hperbolic identities. The proofs of those not lred obtined re left s eercises theorem cosh + sinh = e sinh( + ) = sinh cosh + cosh sinh + = (cos t, sin t) cosh sinh = e cosh( + ) = cosh cosh + sinh sinh cosh sinh = sinh( ) = sinh cosh cosh sinh tnh = sech cosh( ) = cosh cosh sinh sinh coth = csch sinh = sinh cosh cosh( ) = cosh cosh = cosh + sinh sinh( ) = sinh cosh = sinh + = cosh () = WHY THEY ARE CALLED HYPERBOLIC FUNCTIONS Recll tht the prmetric equtions Figure (b) (cosh t, sinh t) = cos t, = sin t ( t π) represent the unit circle + = (Figure 6.9.3), s m be seen b writing + = cos t + sin t = If t π, then the prmeter t cn be interpreted s the ngle in rdins from the positive -is to the point (cos t, sin t) or, lterntivel, s twice the shded re of the sector in Figure (verif). Anlogousl, the prmetric equtions = cosh t, = sinh t ( <t<+ )

65 6.9 Hperbolic Functions nd Hnging Cbles 477 represent portion of the curve =, s m be seen b writing = cosh t sinh t = nd observing tht = cosh t>. This curve, which is shown in Figure 6.9.3b, is the right hlf of lrger curve clled the unit hperbol; this is the reson wh the functions in this section re clled hperbolic functions. It cn be shown tht if t, then the prmeter t cn be interpreted s twice the shded re in Figure 6.9.3b. (We omit the detils.) DERIVATIVE AND INTEGRAL FORMULAS Derivtive formuls for sinh nd cosh cn be obtined b epressing these functions in terms of e nd e : [ d d e e ] [sinh ] = = e + e = cosh d d [ d d e + e ] [cosh ] = = e e = sinh d d Derivtives of the remining hperbolic functions cn be obtined b epressing them in terms of sinh nd cosh nd ppling pproprite identities. For emple, [ ] d d sinh cosh d d [sinh ] sinh [cosh ] [tnh ] = = d d d d cosh cosh = cosh sinh cosh = cosh = sech The following theorem provides complete list of the generlized derivtive formuls nd corresponding integrtion formuls for the hperbolic functions theorem d du [sinh u] =cosh u d d d du [cosh u] =sinh u d d d d [tnh u] =sech u du d d d [coth u] = csch u du d d du [sech u] = sech u tnh u d d d du [csch u] = csch u coth u d d cosh udu= sinh u + C sinh udu= cosh u + C sech udu= tnh u + C csch udu= coth u + C sech u tnh udu= sech u + C csch u coth udu= csch u + C Emple d d [cosh(3 )]=sinh( 3 d ) d [3 ]=3 sinh( 3 ) d [ln(tnh )]= d tnh d d [tnh ] =sech tnh

66 478 Chpter 6 / Applictions of the Definite Integrl in Geometr, Science, nd Engineering Emple 3 sinh 5 cosh d= 6 sinh6 + C tnh d= sinh cosh d = ln cosh +C u = cosh du = sinh d = ln(cosh ) + C u = sinh du = cosh d We were justified in dropping the bsolute vlue signs since cosh > for ll. Emple 4 A ft wire is ttched t its ends to the tops of two 5 ft poles tht re positioned 9 ft prt. How high bove the ground is the middle of the wire? Solution. From bove, the wire forms ctenr curve with eqution ( ) = cosh + c = 56. cosh Figure where the origin is on the ground midw between the poles. Using Formul (4) of Section 6.4 for the length of the ctenr, we hve 45 ( ) d = + d 45 d 45 ( ) d B smmetr = + d d bout the -is = = ( + sinh ( ) cosh d ) d B () nd the fct tht cosh > ( ) ] 45 ( ) 45 = sinh = sinh Using clculting utilit s numeric solver to solve ( ) 45 = sinh for gives 56.. Then 5 = (45) = 56. cosh ( ) 45 + c c 56. so c 5.8. Thus, the middle of the wire is () = 3.93 ft bove the ground (Figure 6.9.4). INVERSES OF HYPERBOLIC FUNCTIONS Referring to Figure 6.9., it is evident tht the grphs of sinh, tnh, coth, nd csch pss the horizontl line test, but the grphs of cosh nd sech do not. In the ltter cse, restricting to be nonnegtive mkes the functions invertible (Figure 6.9.5). The grphs of the si inverse hperbolic functions in Figure were obtined b reflecting the grphs of the hperbolic functions (with the pproprite restrictions) bout the line =.

67 6.9 Hperbolic Functions nd Hnging Cbles 479 Tble 6.9. summrizes the bsic properties of the inverse hperbolic functions. You should confirm tht the domins nd rnges listed in this tble gree with the grphs in Figure = cosh = sech With the restriction tht, the curves = cosh nd = sech pss the horizontl line test. Figure = sinh = cosh = tnh Figure = coth = sech = csch function Tble 6.9. properties of inverse hperbolic functions domin rnge bsic reltionships sinh (, + ) (, + ) sinh (sinh ) = if < < + sinh(sinh ) = if < < + cosh [, + ) [, + ) cosh (cosh ) = if cosh(cosh ) = if tnh (, ) (, + ) tnh (tnh ) = if < < + tnh(tnh ) = if < < coth (, ) (, + ) (, ) (, + ) coth (coth ) = if < or > coth(coth ) = if < or > sech (, ] [, + ) sech (sech ) = if sech(sech ) = if < csch (, ) (, + ) (, ) (, + ) csch (csch ) = if < or > csch(csch ) = if < or >

68 48 Chpter 6 / Applictions of the Definite Integrl in Geometr, Science, nd Engineering LOGARITHMIC FORMS OF INVERSE HYPERBOLIC FUNCTIONS Becuse the hperbolic functions re epressible in terms of e, it should not be surprising tht the inverse hperbolic functions re epressible in terms of nturl logrithms; the net theorem shows tht this is so theorem The following reltionships hold for ll in the domins of the stted inverse hperbolic functions: sinh = ln( + + ) cosh = ln( + ) tnh = ( ) + ln coth = ( ) + ln ( sech + ) ( ) = ln csch + = ln + We will show how to derive the first formul in this theorem nd leve the rest s eercises. The bsic ide is to write the eqution = sinh in terms of eponentil functions nd solve this eqution for s function of. This will produce the eqution = sinh with sinh epressed in terms of nturl logrithms. Epressing = sinh in terms of eponentils ields = sinh = e e which cn be rewritten s e e = Multipling this eqution through b e we obtin nd ppling the qudrtic formul ields e e = e = ± = ± + Since e >, the solution involving the minus sign is etrneous nd must be discrded. Thus, e = + + Tking nturl logrithms ields = ln( + + ) or sinh = ln( + + ) Emple 5 sinh = ln( + + ) = ln( + ).884 ( ) ( ) tnh = + ln = ln

69 6.9 Hperbolic Functions nd Hnging Cbles 48 Show tht the derivtive of the function sinh cn lso be obtined b letting = sinh nd then differentiting = sinh implicitl. DERIVATIVES AND INTEGRALS INVOLVING INVERSE HYPERBOLIC FUNCTIONS Formuls for the derivtives of the inverse hperbolic functions cn be obtined from Theorem For emple, d d [sinh ]= d d [ln( + ( ) + )] = = ( + + )( + ) = + This computtion leds to two integrl formuls, formul tht involves sinh nd n equivlent formul tht involves logrithms: d + = sinh + C = ln( + + ) + C The following two theorems list the generlized derivtive formuls nd corresponding integrtion formuls for the inverse hperbolic functions. Some of the proofs pper s eercises theorem d du d (sinh u) = + u d d du d (cosh u) = u d d (coth u) = du u d, u > d, u > d d (sech u) = u du u d, <u< d d (tnh u) = du u d, u < d d (csch u) = u du + u d, u = theorem If >, then du ( = sinh u ) + C or ln(u + u + u + ) + C du ( = cosh u ) + C or ln(u + u u ) + C, u> du ( u ) u = tnh + C, u < ( u ) or ln + u u + C, u = coth + C, u > ( du u = u sech u + C or ln + ) u + C, < u < u ( du u = + u csch u + C or ln + ) + u + C, u = u

70 48 Chpter 6 / Applictions of the Definite Integrl in Geometr, Science, nd Engineering Emple 6 Evlute d 4 9, >3. Solution. Let u =. Thus, du = d nd d 4 9 = d 4 9 = du u 3 = ( u ) cosh + C = ( ) 3 cosh + C 3 Alterntivel, we cn use the logrithmic equivlent of cosh (/3), ( ) cosh = ln( ) ln 3 (verif), nd epress the nswer s d 4 9 = ln( ) + C QUICK CHECK EXERCISES 6.9 (See pge 485 for nswers.). cosh = sinh = tnh =. Complete the tble. domin rnge cosh sinh tnh coth sech csch d d [cosh ] = d d [tnh ] = d cosh d= tnh d= [sinh ] = d sinh d= 3. The prmetric equtions = cosh t, = sinh t ( <t<+ ) represent the right hlf of the curve clled. Eliminting the prmeter, the eqution of this curve is. d 6. d [cosh ]= d d [tnh ]= d d [sinh ]= EXERCISE SET 6.9 Grphing Utilit Approimte the epression to four deciml plces.. () sinh 3 (b) cosh( ) (c) tnh(ln 4) (d) sinh ( ) (e) cosh 3 (f ) tnh 3 4. () csch( ) (b) sech(ln ) (c) coth (d) sech (e) coth 3 (f ) csch ( 3 ) 3. Find the ect numericl vlue of ech epression. () sinh(ln 3) (b) cosh( ln ) (c) tnh(ln5) (d) sinh( 3ln) 4. In ech prt, rewrite the epression s rtio of polnomils. () cosh(ln ) (b) sinh(ln ) (c) tnh(ln) (d) cosh( ln ) 5. In ech prt, vlue for one of the hperbolic functions is given t n unspecified positive number. Use pproprite identities to find the ect vlues of the remining five hperbolic functions t. () sinh = (b) cosh = 5 (c) tnh 4 = Obtin the derivtive formuls for csch, sech, nd coth from the derivtive formuls for sinh, cosh, nd tnh. 7. Find the derivtives of cosh nd tnh b differentiting the formuls in Theorem Find the derivtives of sinh, cosh, nd tnh b differentiting the equtions = sinh, = cosh, nd = tnh implicitl. 9 8 Find d/d. 9. = sinh(4 8). = cosh( 4 )

71 6.9 Hperbolic Functions nd Hnging Cbles 483. = coth(ln ). = ln(tnh ) 3. = csch(/) 4. = sech(e ) 5. = 4 + cosh (5) 6. = sinh 3 () 7. = 3 tnh ( ) 8. = sinh(cos 3) 9. = sinh ( 3 ). = sinh (/). = ln(cosh ). = cosh (sinh ) 3. = tnh 4. = (coth ) 5. = cosh (cosh ) 6. = sinh (tnh ) 7. = e sech 8. = ( + csch ) 9 44 Evlute the integrls. 9. sinh 6 cosh d 3. cosh( 3)d tnh 3. sech d 3. csch (3)d 33. tnh d 34. coth csch d ln 3 ln tnh sech 3 d 36. d d ( < ) 4. e d / d 44. ln 3 e e d e + e d ( > ) sin θdθ + cos θ d dt t + ( > 5/3) True Flse Determine whether the sttement is true or flse. Eplin our nswer. 45. The eqution cosh = sinh hs no solutions. 46. Ectl two of the hperbolic functions re bounded. 47. There is ectl one hperbolic function f() such tht for ll rel numbers, the eqution f() = hs unique solution. 48. The identities in Theorem 6.9. m be obtined from the corresponding trigonometric identities b replcing ech trigonometric function with its hperbolic nlogue. 49. Find the re enclosed b = sinh, =, nd = ln Find the volume of the solid tht is generted when the region enclosed b = sech, =, =, nd = ln is revolved bout the -is. 5. Find the volume of the solid tht is generted when the region enclosed b = cosh, = sinh, =, nd = 5 is revolved bout the -is. 5. Approimte the positive vlue of the constnt such tht the re enclosed b = cosh, =, =, nd = is squre units. Epress our nswer to t lest five deciml plces. 53. Find the rc length of the ctenr = cosh between = nd = ln. 54. Find the rc length of the ctenr = cosh(/)between = nd = ( > ). 55. In prts () (f ) find the limits, nd confirm tht the re consistent with the grphs in Figures 6.9. nd () lim sinh (b) lim sinh + (c) tnh (d) lim tnh (e) lim + lim + sinh (f ) lim tnh FOCUS ON CONCEPTS 56. Eplin how to obtin the smptotes for = tnh from the curviliner smptotes for = cosh nd = sinh. 57. Prove tht sinh is n odd function of nd tht cosh is n even function of, nd check tht this is consistent with the grphs in Figure Prove the identities. 58. () cosh + sinh = e (b) cosh sinh = e (c) sinh( + ) = sinh cosh + cosh sinh (d) sinh = sinh cosh (e) cosh( + ) = cosh cosh + sinh sinh (f ) cosh = cosh + sinh (g) cosh = sinh + (h) cosh = cosh 59. () tnh = sech tnh + tnh (b) tnh( + ) = + tnh tnh (c) tnh = tnh + tnh 6. Prove: () cosh = ln( + ), (b) tnh = ( ) + ln, <<. 6. Use Eercise 6 to obtin the derivtive formuls for cosh nd tnh. 6. Prove: sech = cosh (/), < coth = tnh (/), > csch = sinh (/), = 63. Use Eercise 6 to epress the integrl du u entirel in terms of tnh.

72 484 Chpter 6 / Applictions of the Definite Integrl in Geometr, Science, nd Engineering 64. Show tht d () d [sech ]= d (b) d [csch ]= In ech prt, find the limit. cosh e () lim + (cosh ln ) (b) lim Use the first nd second derivtives to show tht the grph of = tnh is lws incresing nd hs n inflection point t the origin. 67. The integrtion formuls for / u in Theorem re vlid for u>. Show tht the following formul is vlid for u< : du ( = cosh u ) + C or ln u + u u + C 68. Show tht (sinh + cosh ) n = sinh n + cosh n. 69. Show tht e t d = sinh t t 7. A cble is suspended between two poles s shown in Figure Assume tht the eqution of the curve formed b the cble is = cosh(/), where is positive constnt. Suppose tht the -coordintes of the points of support re = b nd = b, where b>. () Show tht the length L of the cble is given b L = sinh b (b) Show tht the sg S (the verticl distnce between the highest nd lowest points on the cble) is given b S = cosh b 7 7 These eercises refer to the hnging cble described in Eercise Assuming tht the poles re 4 ft prt nd the sg in the cble is 3 ft, pproimte the length of the cble b pproimting. Epress our finl nswer to the nerest tenth of foot. [Hint: First let u = /.] 7. Assuming tht the cble is ft long nd the poles re ft prt, pproimte the sg in the cble b pproimting. Epress our finl nswer to the nerest tenth of foot. [Hint: First let u = 5/.] 73. The design of the Gtew Arch in St. Louis, Missouri, b rchitect Eero Srinn ws implemented using equtions provided b Dr. Hnnskrl Bdel. The eqution used for the centerline of the rch ws = cosh(.333) ft for between nd () Use grphing utilit to grph the centerline of the rch. (b) Find the length of the centerline to four deciml plces. (c) For wht vlues of is the height of the rch ft? Round our nswers to four deciml plces. (d) Approimte, to the nerest degree, the cute ngle tht the tngent line to the centerline mkes with the ground t the ends of the rch. 74. Suppose tht hollow tube rottes with constnt ngulr velocit of ω rd/s bout horizontl is t one end of the tube, s shown in the ccompning figure. Assume tht n object is free to slide without friction in the tube while the tube is rotting. Let r be the distnce from the object to the pivot point t time t, nd ssume tht the object is t rest nd r = when t =. It cn be shown tht if the tube is horizontl t time t = nd rotting s shown in the figure, then r = g [sinh(ωt) sin(ωt)] ω during the period tht the object is in the tube. Assume tht t is in seconds nd r is in meters, nd use g = 9.8m/s nd ω = rd/s. () Grph r versus t for t. (b) Assuming tht the tube hs length of m, pproimtel how long does it tke for the object to rech the end of the tube? (c) Use the result of prt (b) to pproimte dr/dt t the instnt tht the object reches the end of the tube. v r Figure E The ccompning figure (on the net pge) shows person pulling bot b holding rope of length ttched to the bow nd wlking long the edge of dock. If we ssume tht the rope is lws tngent to the curve trced b the bow of the bot, then this curve, which is clled trctri, hs the propert tht the segment of the tngent line between the curve nd the -is hs constnt length. It cn be proved tht the eqution of this trctri is = sech () Show tht to move the bow of the bot to point (, ), the person must wlk distnce D = sech from the origin. (b) If the rope hs length of 5 m, how fr must the person wlk from the origin to bring the bot m from the dock? Round our nswer to two deciml plces. (c) Find the distnce trveled b the bow long the trctri s it moves from its initil position to the point where it is 5 m from the dock.

73 Chpter 6 Review Eercises 485 Dock (, ) (, ) Initil position Figure E Writing Suppose tht, b nlog with the trigonometric functions, we define cosh t nd sinh t geometricll using Figure 6.9.3b: For n rel number t, define = cosh t nd = sinh t to be the unique vlues of nd such tht (i) P(,) is on the right brnch of the unit hperbol = ; (ii) t nd hve the sme sign (or re both ); (iii) the re of the region bounded b the -is, the right brnch of the unit hperbol, nd the segment from the origin to P is t /. Discuss wht properties would first need to be verified in order for this to be legitimte definition. 77. Writing Investigte wht properties of cosh t nd sinh t cn be proved directl from the geometric definition in Eercise 76. Write short description of the results of our investigtion. QUICK CHECK ANSWERS e + e ; domin rnge e e ; e e e + e cosh sinh tnh coth sech csch (, + ) (, + ) (, + ) (, ) (, + ) (, + ) (, ) (, + ) [, + ) (, + ) (, ) (, ) (, + ) (, ] (, ) (, + ) 3. unit hperbol; = 4. sinh ; cosh ; sech 5. sinh + C; cosh + C; ln(cosh ) + C 6. ; ; + CHAPTER 6 REVIEW EXERCISES. Describe the method of slicing for finding volumes, nd use tht method to derive n integrl formul for finding volumes b the method of disks.. Stte n integrl formul for finding volume b the method of clindricl shells, nd use Riemnn sums to derive the formul. 3. Stte n integrl formul for finding the rc length of smooth curve = f()over n intervl [,b], nd use Riemnn sums to derive the formul. 4. Stte n integrl formul for the work W done b vrible force F() pplied in the direction of motion to n object moving from = to = b, nd use Riemnn sums to derive the formul. 5. Stte n integrl formul for the fluid force F eerted on verticl flt surfce immersed in fluid of weight densit ρ, nd use Riemnn sums to derive the formul. 6. Let R be the region in the first qudrnt enclosed b =, = +, nd =. In ech prt, set up, but do not evlute, n integrl or sum of integrls tht will solve the problem. () Find the re of R b integrting with respect to. (b) Find the re of R b integrting with respect to. (c) Find the volume of the solid generted b revolving R bout the -is b integrting with respect to. (d) Find the volume of the solid generted b revolving R bout the -is b integrting with respect to. (e) Find the volume of the solid generted b revolving R bout the -is b integrting with respect to. (f ) Find the volume of the solid generted b revolving R bout the -is b integrting with respect to. (g) Find the volume of the solid generted b revolving R bout the line = 3b integrting with respect to. (h) Find the volume of the solid generted b revolving R bout the line = 5 b integrting with respect to. 7. () Set up sum of definite integrls tht represents the totl shded re between the curves = f()nd = g() in the ccompning figure on the net pge. (cont.)

74 486 Chpter 6 / Applictions of the Definite Integrl in Geometr, Science, nd Engineering (b) Find the totl re enclosed between = 3 nd = over the intervl [, ]. = f() b c d = g() Figure E-7 8. The ccompning figure shows velocit versus time curves for two crs tht move long stright trck, ccelerting from rest t common strting line. () How fr prt re the crs fter 6 seconds? (b) How fr prt re the crs fter T seconds, where T 6? 8 v (ft/s) v (t) = 3t v (t) = t / t (s) 6 Figure E-8 9. Let R be the region enclosed b the curves = + 4, = 3, nd the -is. Find nd evlute definite integrl tht represents the volume of the solid generted b revolving R bout the -is.. A footbll hs the shpe of the solid generted b revolving the region bounded between the -is nd the prbol = 4R( 4 L )/L bout the -is. Find its volume.. Find the volume of the solid whose bse is the region bounded between the curves = nd = / for 4 nd whose cross sections perpendiculr to the -is re squres.. Consider the region enclosed b = sin, =, nd =. Set up, but do not evlute, n integrl tht represents the volume of the solid generted b revolving the region bout the -is using () disks (b) clindricl shells. 3. Find the rc length in the second qudrnt of the curve /3 + /3 = 4 from = 8to =. 4. Let C be the curve = e between = nd = ln. In ech prt, set up, but do not evlute, n integrl tht solves the problem. () Find the rc length of C b integrting with respect to. (b) Find the rc length of C b integrting with respect to. 5. Find the re of the surfce generted b revolving the curve = 5,9 6, bout the -is. 6. Let C be the curve 7 3 = between = nd =. In ech prt, set up, but do not evlute, n integrl or sum of integrls tht solves the problem. () Find the re of the surfce generted b revolving C bout the -is b integrting with respect to. (b) Find the re of the surfce generted b revolving C bout the -is b integrting with respect to. (c) Find the re of the surfce generted b revolving C bout the line = b integrting with respect to. 7. Consider the solid generted b revolving the region enclosed b = sec, =, = π/3, nd = bout the -is. Find the verge vlue of the re of cross section of this solid tken perpendiculr to the -is. 8. Consider the solid generted b revolving the region enclosed b = nd = bout the -is. Without performing n integrtion, find the verge vlue of the re of cross section of this solid tken perpendiculr to the -is. 9. () A spring eerts force of.5 N when stretched.5 m beond its nturl length. Assuming tht Hooke s lw pplies, how much work ws performed in stretching the spring to this length? (b) How fr beond its nturl length cn the spring be stretched with 5 J of work?. A bot is nchored so tht the nchor is 5 ft below the surfce of the wter. In the wter, the nchor weighs lb nd the chin weighs 3 lb/ft. How much work is required to rise the nchor to the surfce? Find the centroid of the region.. The region bounded b = 4 nd = 8( ).. The upper hlf of the ellipse (/) + (/b) =. 3. In ech prt, set up, but do not evlute, n integrl tht solves the problem. () Find the fluid force eerted on side of bo tht hs 3 m squre bse nd is filled to depth of m with liquid of weight densit ρ N/m 3. (b) Find the fluid force eerted b liquid of weight densit ρ lb/ft 3 on fce of the verticl plte shown in prt () of the ccompning figure. (c) Find the fluid force eerted on the prbolic dm in prt (b) of the ccompning figure b wter tht etends to the top of the dm. ft ft 4 ft () Figure E-3 5 m m 4. Show tht for n constnt, the function = sinh() stisfies the eqution =. 5. In ech prt, prove the identit. () cosh 3 = 4 cosh 3 3 cosh (b) cosh = (cosh + ) (c) sinh =± (cosh ) (b)

75 Chpter 6 Mking Connections 487 CHAPTER 6 MAKING CONNECTIONS. Suppose tht f is nonnegtive function defined on [, ] such tht the re between the grph of f nd the intervl [, ] is A nd such tht the re of the region R between the grph of g() = f( ) nd the intervl [, ] is A. In ech prt, epress our nswer in terms of A nd A. () Wht is the volume of the solid of revolution generted b revolving R bout the -is? (b) Find vlue of such tht if the -plne were horizontl, the region R would blnce on the line =.. A wter tnk hs the shpe of conicl frustum with rdius of the bse 5 ft, rdius of the top ft nd (verticl) height 5 ft. Suppose the tnk is filled with wter nd consider the problem of finding the work required to pump ll the wter out through hole in the top of the tnk. () Solve this problem using the method of Emple 5 in Section 6.6. (b) Solve this problem using Definition [Hint: Think of the bse s the hed of piston tht epnds to wtertight fit ginst the sides of the tnk s the piston is pushed upwrd. Wht importnt result bout wter pressure do ou need to use?] 3. A disk of rdius is n inhomogeneous lmin whose densit is function f(r) of the distnce r to the center of the lmin. Modif the rgument used to derive the method of clindricl shells to find formul for the mss of the lmin. 4. Compre Formul () in Section 6.7 with Formul (8) in Section 6.8. Then give plusible rgument tht the force on flt surfce immersed verticll in fluid of constnt weight densit is equl to the product of the re of the surfce nd the pressure t the centroid of the surfce. Conclude tht the force on the surfce is the sme s if the surfce were immersed horizontll t the depth of the centroid. 5. Archimedes Principle sttes tht solid immersed in fluid eperiences buont force equl to the weight of the fluid displced b the solid. () Use the results of Section 6.8 to verif Archimedes Principle in the cse of (i) bo-shped solid with pir of fces prllel to the surfce of the fluid, (ii) solid clinder with verticl is, nd (iii) clindricl shell with verticl is. (b) Give plusible rgument for Archimedes Principle in the cse of solid of revolution immersed in fluid such tht the is of revolution of the solid is verticl. [Hint: Approimte the solid b union of clindricl shells nd use the result from prt ().]

76 7 PRINCIPLES OF INTEGRAL EVALUATION AP/Wide World Photos The floting roof on the Stde de Frnce sports comple is n ellipse. Finding the rc length of n ellipse involves numericl integrtion techniques introduced in this chpter. In erlier chpters we obtined mn bsic integrtion formuls s n immedite consequence of the corresponding differentition formuls. For emple, knowing tht the derivtive of sin is cos enbled us to deduce tht the integrl of cos is sin. Subsequentl, we epnded our integrtion repertoire b introducing the method of u-substitution. Tht method enbled us to integrte mn functions b trnsforming the integrnd of n unfmilir integrl into fmilir form. However, u-substitution lone is not dequte to hndle the wide vriet of integrls tht rise in pplictions, so dditionl integrtion techniques re still needed. In this chpter we will discuss some of those techniques, nd we will provide more sstemtic procedure for ttcking unfmilir integrls. We will tlk more bout numericl pproimtions of definite integrls, nd we will eplore the ide of integrting over infinite intervls. 7. AN OVERVIEW OF INTEGRATION METHODS In this section we will give brief overview of methods for evluting integrls, nd we will review the integrtion formuls tht were discussed in erlier sections. 488 METHODS FOR APPROACHING INTEGRATION PROBLEMS There re three bsic pproches for evluting unfmilir integrls: Technolog CAS progrms such s Mthemtic, Mple, nd the open source progrm Sge re cpble of evluting etremel complicted integrls, nd such progrms re incresingl vilble for both computers nd hndheld clcultors. Tbles Prior to the development of CAS progrms, scientists relied hevil on tbles to evlute difficult integrls rising in pplictions. Such tbles were compiled over mn ers, incorporting the skills nd eperience of mn people. One such tble ppers in the endppers of this tet, but more comprehensive tbles pper in vrious reference books such s the CRC Stndrd Mthemticl Tbles nd Formule, CRC Press, Inc.,. Trnsformtion Methods Trnsformtion methods re methods for converting unfmilir integrls into fmilir integrls. These include u-substitution, lgebric mnipultion of the integrnd, nd other methods tht we will discuss in this chpter.

77 7. An Overview of Integrtion Methods 489 None of the three methods is perfect; for emple, CAS progrms often encounter integrls tht the cnnot evlute nd the sometimes produce nswers tht re unnecessril complicted, tbles re not ehustive nd m not include prticulr integrl of interest, nd trnsformtion methods rel on humn ingenuit tht m prove to be indequte in difficult problems. In this chpter we will focus on trnsformtion methods nd tbles, so it will not be necessr to hve CAS such s Mthemtic, Mple, or Sge. However, if ou hve CAS, then ou cn use it to confirm the results in the emples, nd there re eercises tht re designed to be solved with CAS. If ou hve CAS, keep in mind tht mn of the lgorithms tht it uses re bsed on the methods we will discuss here, so n understnding of these methods will help ou to use our technolog in more informed w. A REVIEW OF FAMILIAR INTEGRATION FORMULAS The following is list of bsic integrls tht we hve encountered thus fr: CONSTANTS, POWERS, EXPONENTIALS. du = u + C. du= du = u + C 3. u r du = ur+ du + C, r 4. = ln u +C r + u 5. e u du = e u + C 6. b u du = bu + C, b>,b = ln b TRIGONOMETRIC FUNCTIONS 7. sin udu= cos u + C 8. cos udu= sin u + C 9. sec udu= tn u + C. csc udu= cot u + C. sec u tn udu= sec u + C. csc u cot udu= csc u + C 3. tn udu= ln cos u +C 4. cot udu= ln sin u +C HYPERBOLIC FUNCTIONS 5. sinh udu= cosh u + C sech udu= tnh u + C sech u tnh udu= sech u + C. cosh udu= sinh u + C csch udu= coth u + C csch u coth udu= csch u + C ALGEBRAIC FUNCTIONS ( > ) du. = u u sin + C ( u <) du. + u = u tn + C du 3. = u u sec u + C ( << u )

78 49 Chpter 7 / Principles of Integrl Evlution du 4. = ln(u + u + ) + C + u 5. du u = ln u + u + C du 6. u = ln + u u + C du 7. u = u ln + u u + C du 8. u = + u ln + + u u + C ( << u ) ( < u <) REMARK Formul 5 is generliztion of result in Theorem Reders who did not cover Section 6.9 cn ignore Formuls 4 8 for now, since we will develop other methods for obtining them in this chpter. QUICK CHECK EXERCISES 7. (See pge 49 for nswers.). Use lgebric mnipultion nd (if necessr) u-substitution (c) (cot + )d = to integrte the function. + () d = (d) sec + tn d = + (b) + d = 3. Integrte the function. () d = + (c) + d = (b) e + d = (d) e 3ln d = (c) (sin 3 cos + sin cos 3 )d =. Use trigonometric identities nd (if necessr) u-substitution to integrte the function. (d) () csc d = (e + e ) d = (b) cos d = EXERCISE SET 7. 3 Evlute the integrls b mking pproprite u-substitutions nd ppling the formuls reviewed in this section.. (4 ) 3 d d 3. sec ( )d 4. 4 tn( )d sin 3 5. d 6. + cos d sec(ln )tn(ln ) 7. e sinh(e )d 8. d 9. e tn sec d. d cos 5 cos 5 sin 5d. d sin sin + e e tn d e + d e d 6. ( + ) cot( + )d cosh d 8. d 3 d (ln )

79 sec(sin θ)tn(sin θ)cos θdθ csch (/) d. e d 4. 4 e e d 6. e d 8. csc( ) 4 d 3. d 4 cos(ln ) d sinh( / ) d 3/ e d 4 e π d 3. () Derive the identit 7. Integrtion b Prts 49 sech + tnh = sech (b) Use the result in prt () to evlute sech d. (c) Derive the identit sech = e e + (d) Use the result in prt (c) to evlute sech d. (e) Eplin wh our nswers to prts (b) nd (d) re consistent. 33. () Derive the identit FOCUS ON CONCEPTS 3. () Evlute the integrl sin cos dusing the substitution u = sin. (b) Evlute the integrl sin cos dusing the identit sin = sin cos. (c) Eplin wh our nswers to prts () nd (b) re consistent. sec tn = sin cos (b) Use the identit sin = sin cos long with the result in prt () to evlute csc d. (c) Use the identit cos = sin[(π/) ] long with our nswer to prt () to evlute sec d. QUICK CHECK ANSWERS 7.. () + ln +C (b) + ln + +C (c) ln( + ) + tn + C (d) 5 + C. () cos + C (b) tn + C 5 (c) cot + C (d) ln( + sin ) + C 3. () 3 ( )3/ + C (b) e+ + C (c) sin + C (d) 4 tnh + C 7. INTEGRATION BY PARTS In this section we will discuss n integrtion technique tht is essentill n ntiderivtive formultion of the formul for differentiting product of two functions. THE PRODUCT RULE AND INTEGRATION BY PARTS Our primr gol in this section is to develop generl method for ttcking integrls of the form f()g()d As first step, let G() be n ntiderivtive of g(). In this cse G () = g(), so the product rule for differentiting f()g() cn be epressed s d d [f()g()] =f()g () + f ()G() = f()g() + f ()G() () This implies tht f()g() is n ntiderivtive of the function on the right side of (), so we cn epress () in integrl form s [f()g() + f ()G()] d = f()g()

80 49 Chpter 7 / Principles of Integrl Evlution or, equivlentl, s f()g()d = f()g() f ()G() d () This formul llows us to integrte f()g()b integrting f ()G() insted, nd in mn cses the net effect is to replce difficult integrtion with n esier one. The ppliction of this formul is clled integrtion b prts. In prctice, we usull rewrite () b letting u = f(), du = f () d v = G(), dv = G () d = g() d This ields the following lterntive form for (): udv = uv vdu (3) Note tht in Emple we omitted the constnt of integrtion in clculting v from dv. Hd we included constnt of integrtion, it would hve eventull dropped out. This is lws the cse in integrtion b prts [Eercise 68(b)], so it is common to omit the constnt t this stge of the computtion. However, there re certin cses in which mking clever choice of constnt of integrtion to include with v cn simplif the computtion of vdu (Eercises 69 7). Emple Use integrtion b prts to evlute cos d. Solution. We will ppl Formul (3). The first step is to mke choice for u nd dv to put the given integrl in the form udv. We will let u = nd dv = cos d (Other possibilities will be considered lter.) The second step is to compute dufrom u nd v from dv. This ields du = d nd v = dv = cos d= sin The third step is to ppl Formul (3). This ields }{{} cos }{{ d } = }{{} sin }{{ } sin }{{} }{{} d u dv u v v du = sin ( cos ) + C = sin + cos + C GUIDELINES FOR INTEGRATION BY PARTS The min gol in integrtion b prts is to choose u nd dv to obtin new integrl tht is esier to evlute thn the originl. In generl, there re no hrd nd fst rules for doing this; it is minl mtter of eperience tht comes from lots of prctice. A strteg tht often works is to choose u nd dv so tht u becomes simpler when differentited, while leving dv tht cn be redil integrted to obtin v. Thus, for the integrl cos d in Emple, both gols were chieved b letting u = nd dv = cos d. In contrst, u = cos would not hve been good first choice in tht emple, since du/d = sin is no simpler thn u. Indeed, hd we chosen u = cos dv = d du = sin d v= d= then we would hve obtined cos d= cos ( sin )d = cos + sin d For this choice of u nd dv, the new integrl is ctull more complicted thn the originl.

81 7. Integrtion b Prts 493 The LIATE method is discussed in the rticle A Technique for Integrtion b Prts, Americn Mthemticl Monthl, Vol. 9, 983, pp., b Herbert Ksube. There is nother useful strteg for choosing u nd dv tht cn be pplied when the integrnd is product of two functions from different ctegories in the list Logrithmic, Inverse trigonometric, Algebric, Trigonometric, Eponentil In this cse ou will often be successful if ou tke u to be the function whose ctegor occurs erlier in the list nd tke dv to be the rest of the integrnd. The cronm LIATE will help ou to remember the order. The method does not work ll the time, but it works often enough to be useful. Note, for emple, tht the integrnd in Emple consists of the product of the lgebric function nd the trigonometric function cos. Thus, the LIATE method suggests tht we should let u = nd dv = cos d, which proved to be successful choice. Emple Evlute e d. Solution. In this cse the integrnd is the product of the lgebric function with the eponentil function e. According to LIATE we should let so tht u = nd dv = e d du = d nd v = e d = e Thus, from (3) e d = udv = uv vdu= e e d = e e + C Emple 3 Evlute ln d. Solution. One choice is to let u = nd dv = ln d. But with this choice finding v is equivlent to evluting ln d nd we hve gined nothing. Therefore, the onl resonble choice is to let u = ln dv = d du = d v = d = With this choice it follows from (3) tht ln d= udv = uv vdu= ln d = ln + C REPEATED INTEGRATION BY PARTS It is sometimes necessr to use integrtion b prts more thn once in the sme problem. Emple 4 Evlute e d. Solution. Let u =, dv = e d, du = d, v= e d = e

82 494 Chpter 7 / Principles of Integrl Evlution so tht from (3) e d = udv = uv vdu e ()d e d (4) = ( e ) = e + The lst integrl is similr to the originl ecept tht we hve replced b. Another integrtion b prts pplied to e d will complete the problem. We let u =, dv = e d, du = d, v = e d = e so tht e d = ( e ) e d = e + e d = e e + C Finll, substituting this into the lst line of (4) ields e d = e + e d = e + ( e e ) + C = ( + + )e + C The LIATE method suggests tht integrls of the form e sin b d nd e cos b d cn be evluted b letting u = sin b or u = cos b nd dv = e d. However, this will require technique tht deserves specil ttention. Emple 5 Evlute e cos d. Solution. Let u = cos, dv = e d, du = sin d, v= e d = e Thus, e cos d= udv = uv vdu= e cos + e sin d (5) Since the integrl e sin d is similr in form to the originl integrl e cos d,it seems tht nothing hs been ccomplished. However, let us integrte this new integrl b prts. We let u = sin, dv = e d, du = cos d, v= e d = e Thus, e sin d= udv = uv vdu= e sin e cos d Together with Eqution (5) this ields e cos d= e cos + e sin e cos d (6)

83 7. Integrtion b Prts 495 which is n eqution we cn solve for the unknown integrl. We obtin e cos d= e cos + e sin nd hence e cos d= e cos + e sin + C More informtion on tbulr integrtion b prts cn be found in the rticles Tbulr Integrtion b Prts, College Mthemtics Journl, Vol., 99, pp. 37 3, b Dvid Horowitz nd More on Tbulr Integrtion b Prts, College Mthemtics Journl, Vol., 99, pp. 47 4, b Leonrd Gillmn. A TABULAR METHOD FOR REPEATED INTEGRATION BY PARTS Integrls of the form p()f()d where p() is polnomil, cn sometimes be evluted using repeted integrtion b prts in which u is tken to be p() or one of its derivtives t ech stge. Since du is computed b differentiting u, the repeted differentition of p() will eventull produce, t which point ou m be left with simplified integrtion problem. A convenient method for orgnizing the computtions into two columns is clled tbulr integrtion b prts. Tbulr Integrtion b Prts Step. Differentite p() repetedl until ou obtin, nd list the results in the first column. Step. Integrte f() repetedl nd list the results in the second column. Step 3. Drw n rrow from ech entr in the first column to the entr tht is one row down in the second column. Step 4. Lbel the rrows with lternting + nd signs, strting with +. Step 5. For ech rrow, form the product of the epressions t its tip nd til nd then multipl tht product b + or in ccordnce with the sign on the rrow. Add the results to obtin the vlue of the integrl. This process is illustrted in Figure 7.. for the integrl ( )cos d. repeted differentition repeted integrtion + + cos sin cos sin Figure 7.. ( ) cos d = ( ) sin + ( ) cos sin + C = ( ) sin + ( ) cos + C Emple 6 In Emple of Section 5.3 we evluted d using u-substitution. Evlute this integrl using tbulr integrtion b prts.

84 496 Chpter 7 / Principles of Integrl Evlution Solution. repeted differentition repeted integrtion + + ( ) / _ ( ) 3/ 3 4 ( ) 5 5/ 8 ( ) 7/ 5 The result obtined in Emple 6 looks quite different from tht obtined in Emple of Section 5.3. Show tht the two nswers re equivlent. Thus, it follows tht d = 3 ( ) 3/ 8 5 ( )5/ ( )7/ + C INTEGRATION BY PARTS FOR DEFINITE INTEGRALS For definite integrls the formul corresponding to (3) is b ] b udv = uv b vdu (7) REMARK It is importnt to keep in mind tht the vribles u nd v in this formul re functions of nd tht the limits of integrtion in (7) re limits on the vrible. Sometimes it is helpful to emphsize this b writing (7) s b ] b b udv = uv vdu (8) = = = The net emple illustrtes how integrtion b prts cn be used to integrte the inverse trigonometric functions. Emple 7 Solution. Let Evlute tn d. u = tn, dv = d, du = + d, v = Thus, tn d= = tn ] udv = uv ] vdu + d The limits of integrtion refer to ; tht is, = nd =. But so + d = + d = ] ln( + ) = ln ] tn d= tn ( π ) ln = 4 ln = π 4 ln

85 7. Integrtion b Prts 497 REDUCTION FORMULAS Integrtion b prts cn be used to derive reduction formuls for integrls. These re formuls tht epress n integrl involving power of function in terms of n integrl tht involves lower power of tht function. For emple, if n is positive integer nd n, then integrtion b prts cn be used to obtin the reduction formuls sin n d= n sinn cos + n sin n d (9) n cos n d= n cosn sin + n n cos n d () To illustrte how such formuls cn be obtined, let us derive (). We begin b writing cos n s cos n cos nd letting so tht u = cos n du = (n ) cos n ( sin )d = (n ) cos n sin d cos n d= cos n cos d= = cos n sin + (n ) = cos n sin + (n ) = cos n sin + (n ) dv = cos d v = sin udv = uv vdu sin cos n d ( cos )cos n d cos n d (n ) cos n d Moving the lst term on the right to the left side ields n cos n d= cos n sin + (n ) cos n d from which () follows. The derivtion of reduction formul (9) is similr (Eercise 63). Reduction formuls (9) nd () reduce the eponent of sine (or cosine) b. Thus, if the formuls re pplied repetedl, the eponent cn eventull be reduced to if n is even or if n is odd, t which point the integrtion cn be completed. We will discuss this method in more detil in the net section, but for now, here is n emple tht illustrtes how reduction formuls work. Emple 8 Evlute cos 4 d. Solution. From () with n = 4 cos 4 d= 4 cos3 sin ( = 4 cos3 sin cos sin + cos d Now ppl () with n =. = 4 cos3 sin cos sin C ) d

86 498 Chpter 7 / Principles of Integrl Evlution QUICK CHECK EXERCISES 7. (See pge 5 for nswers.). () If G () = g(), then f()g()d = f()g() (c) (d) sin d; u =, dv = d; u =, dv = (b) If u = f() nd v = G(), then the formul in prt () cn be written in the form udv =.. Find n pproprite choice of u nd dv for integrtion b prts of ech integrl. Do not evlute the integrl. () ln d; u =, dv = (b) ( ) sin d; u =, dv = 3. Useintegrtion b prts to evlute the integrl. () e d (b) ln( )d π/6 (c) sin 3d 4. Use reduction formul to evlute sin 3 d. EXERCISE SET Evlute the integrl.. e d. 3. e d sin 3d cos d ln d.. (ln ) d. 3. ln(3 )d sin d tn (3)d e sin d.. sin(ln )d. 3. sec d e d e e d 8. ln d 3. ln( + )d 3. e 3 d e d cos d sin d ln d ln d ln( + 4)d cos ()d tn d e 3 cos d cos(ln )d tn d e ( + ) d e 5 d e e ln 3/ d sin d π 3 sec θdθ 34. sin d 36. tn d 38. π sec d ( + cos )d ln( + )d 39 4 True Flse Determine whether the sttement is true or flse. Eplin our nswer. 39. The min gol in integrtion b prts is to choose u nd dv to obtin new integrl tht is esier to evlute thn the originl. 4. Appling the LIATE strteg to evlute 3 ln d,we should choose u = 3 nd dv = ln d. 4. To evlute ln e d using integrtion b prts, choose dv = e d. 4. Tbulr integrtion b prts is useful for integrls of the form p()f()d, where p() is polnomil nd f() cn be repetedl integrted Evlute the integrl b mking u-substitution nd then integrting b prts. 43. e d 44. cos d 45. Prove tht tbulr integrtion b prts gives the correct nswer for p()f()d where p() is n qudrtic polnomil nd f() is n function tht cn be repetedl integrted. 46. The computtions of n integrl evluted b repeted integrtion b prts cn be orgnized using tbulr integrtion b prts. Use this orgniztion to evlute e cos din

87 7. Integrtion b Prts 499 two ws: first b repeted differentition of cos (compre Emple 5), nd then b repeted differentition of e Evlute the integrl using tbulr integrtion b prts. 47. (3 + )e d 48. ( + + ) sin d sin d d 5. e sin b d 5. e 3θ sin 5θ dθ 53. Consider the integrl sin cos d. () Evlute the integrl two ws: first using integrtion b prts, nd then using the substitution u = sin. (b) Show tht the results of prt () re equivlent. (c) Which of the two methods do ou prefer? Discuss the resons for our preference. 54. Evlute the integrl 3 + d using () integrtion b prts (b) the substitution u = () Find the re of the region enclosed b = ln, the line = e, nd the -is. (b) Find the volume of the solid generted when the region in prt () is revolved bout the -is. 56. Find the re of the region between = sin nd = for π/. 57. Find the volume of the solid generted when the region between = sin nd = for π is revolved bout the -is. 58. Find the volume of the solid generted when the region enclosed between = cos nd = for π/ is revolved bout the -is. 59. A prticle moving long the -is hs velocit function v(t) = t 3 sin t. How fr does the prticle trvel from time t = tot = π? 6. The stud of swtooth wves in electricl engineering leds to integrls of the form π/ω π/ω t sin(kωt) dt where k is n integer nd ω is nonzero constnt. Evlute the integrl. 6. Use reduction formul (9) to evlute π/ () sin 4 d (b) sin 5 d. 6. Use reduction formul () to evlute π/ () cos 5 d (b) cos 6 d. 63. Derive reduction formul (9). 64. In ech prt, use integrtion b prts or other methods to derive the reduction formul. () sec n d= secn tn + n sec n d n n (b) tn n d= tnn n tn n d (c) n e d = n e n n e d Use the reduction formuls in Eercise 64 to evlute the integrls. 65. () tn 4 d (b) sec 4 d (c) 3 e d 66. () e 3 d (b) e d [Hint: First mke substitution.] 67. Let f be function whose second derivtive is continuous on [, ]. Show tht f () d = f () + f ( ) f() + f( ) FOCUS ON CONCEPTS 68. () In the integrl cos d, let u =, dv = cos d, du = d, v = sin + C Show tht the constnt C cncels out, thus giving the sme solution obtined b omitting C. (b) Show tht in generl uv vdu= u(v + C ) (v + C )du thereb justifing the omission of the constnt of integrtion when clculting v in integrtion b prts. 69. Evlute ln( + )d using integrtion b prts. Simplif the computtion of vdu b introducing constnt of integrtion C = when going from dv to v. 7. Evlute ln(3 )d using integrtion b prts. Simplif the computtion of vdu b introducing constnt of integrtion C = when going from dv 3 to v. Compre our solution with our nswer to Eercise Evlute tn dusing integrtion b prts. Simplif the computtion of vdub introducing constnt of integrtion C = when going from dv to v. 7. Wht eqution results if integrtion b prts is pplied to the integrl ln d with the choices u = nd dv = ln d? In wht sense is this eqution true? In wht sense is it flse?

88 5 Chpter 7 / Principles of Integrl Evlution 73. Writing Eplin how the product rule for derivtives nd the technique of integrtion b prts re relted. 74. Writing For wht sort of problems re the integrtion techniques of substitution nd integrtion b prts competing techniques? Describe situtions, with emples, where ech of these techniques would be preferred over the other. QUICK CHECK ANSWERS 7.. () f ()G() d (b) uv vdu. () ln ; d (b) ; sin d (c) sin ; d (d) ; ( 3. () ) e + C (b) ( ) ln( ) + C (c) sin cos 3 cos + C d 7.3 INTEGRATING TRIGONOMETRIC FUNCTIONS In the lst section we derived reduction formuls for integrting positive integer powers of sine, cosine, tngent, nd secnt. In this section we will show how to work with those reduction formuls, nd we will discuss methods for integrting other kinds of integrls tht involve trigonometric functions. INTEGRATING POWERS OF SINE AND COSINE We begin b reclling two reduction formuls from the preceding section. sin n d= n sinn cos + n n cos n d= n cosn sin + n n In the cse where n =, these formuls ield sin d= sin cos + cos d= cos sin + sin n d () cos n d () d = sin cos + C (3) d = + sin cos + C (4) Alterntive forms of these integrtion formuls cn be derived from the trigonometric identities sin = ( cos ) nd cos = ( + cos ) (5 6) which follow from the double-ngle formuls cos = sin nd cos = cos These identities ield sin d= ( cos )d = sin + C (7) 4 cos d= ( + cos )d = + sin + C (8) 4

89 7.3 Integrting Trigonometric Functions 5 Observe tht the ntiderivtives in Formuls (3) nd (4) involve both sines nd cosines, wheres those in (7) nd (8) involve sines lone. However, the pprent discrepnc is es to resolve b using the identit sin = sin cos to rewrite (7) nd (8) in forms (3) nd (4), or conversel. In the cse where n = 3, the reduction formuls for integrting sin 3 nd cos 3 ield sin 3 d= 3 sin cos + sin d= 3 3 sin cos cos + C (9) 3 cos 3 d= 3 cos sin + 3 cos d= 3 cos sin + sin + C () 3 TECHNOLOGY MASTERY The Mple CAS produces forms () nd () when sked to integrte sin 3 nd cos 3, but Mthemtic produces sin 3 d= 3 4 cos + cos 3 + C cos 3 d= 3 4 sin + sin 3 + C Use trigonometric identities to reconcile the results of the two progrms. If desired, Formul (9) cn be epressed in terms of cosines lone b using the identit sin = cos, nd Formul () cn be epressed in terms of sines lone b using the identit cos = sin. We leve it for ou to do this nd confirm tht sin 3 d= 3 cos3 cos + C () cos 3 d= sin 3 sin3 + C () We leve it s n eercise to obtin the following formuls b first ppling the reduction formuls, nd then using pproprite trigonometric identities. sin 4 d= 3 8 sin + sin 4 + C (3) 4 3 cos 4 d= sin + sin 4 + C (4) 3 = sin Emple Find the volume V of the solid tht is obtined when the region under the curve = sin over the intervl [,π] is revolved bout the -is (Figure 7.3.). c Solution. ields Using the method of disks, Formul (5) of Section 6., nd Formul (3) bove V = π π sin 4 d= π [ 3 8 sin sin 4] π = 3 8 π Figure 7.3. INTEGRATING PRODUCTS OF SINES AND COSINES If m nd n re positive integers, then the integrl sin m cos n d cn be evluted b one of the three procedures stted in Tble 7.3., depending on whether m nd n re odd or even. Emple Evlute () sin 4 cos 5 d (b) sin 4 cos 4 d

90 5 Chpter 7 / Principles of Integrl Evlution Tble 7.3. integrting products of sines nd cosines sin m cos n d n odd m odd m even n even procedure Split off fctor of cos. Appl the relevnt identit. Mke the substitution u = sin. Split off fctor of sin. Appl the relevnt identit. Mke the substitution u = cos. Use the relevnt identities to reduce the powers on sin nd cos. relevnt identities cos = sin sin = cos sin = ( cos ) cos = ( + cos ) Solution (). Since n = 5 is odd, we will follow the first procedure in Tble 7.3.: sin 4 cos 5 d= sin 4 cos 4 cos d = sin 4 ( sin ) cos d = u 4 ( u ) du = (u 4 u 6 + u 8 )du = 5 u5 7 u7 + 9 u9 + C = 5 sin5 7 sin7 + 9 sin9 + C Solution (b). Since m = n = 4, both eponents re even, so we will follow the third procedure in Tble 7.3.: sin 4 cos 4 d= (sin ) (cos ) d ( = [ cos ]) ( [ + cos ]) d = ( cos ) d 6 sin 4 d = 6 = 3 sin 4 udu Note tht this cn be obtined more directl from the originl integrl using the identit sin cos = sin. u = du = d or d = du = ( u sin u sin 4u) + C Formul (3) = 3 8 sin 4 + sin 8 + C 8 4

91 7.3 Integrting Trigonometric Functions 53 Integrls of the form sin m cos n d, sin m sin n d, cos m cos n d (5) cn be found b using the trigonometric identities sin α cos β = [sin(α β) + sin(α + β)] (6) sin α sin β = [cos(α β) cos(α + β)] (7) cos α cos β = [cos(α β) + cos(α + β)] (8) to epress the integrnd s sum or difference of sines nd cosines. Emple 3 Evlute sin 7 cos 3d. Solution. Using (6) ields sin 7 cos 3d= (sin 4 + sin )d = 8 cos 4 cos + C INTEGRATING POWERS OF TANGENT AND SECANT The procedures for integrting powers of tngent nd secnt closel prllel those for sine nd cosine. The ide is to use the following reduction formuls (which were derived in Eercise 64 of Section 7.) to reduce the eponent in the integrnd until the resulting integrl cn be evluted: tn n d= tnn n sec n d= secn tn n + n n tn n d (9) sec n d () In the cse where n is odd, the eponent cn be reduced to, leving us with the problem of integrting tn or sec. These integrls re given b tn d= ln sec +C () sec d= ln sec + tn +C () Formul () cn be obtined b writing sin tn d= cos d = ln cos +C u = cos du = sin d = ln sec +C ln cos = ln cos To obtin Formul () we write ( ) sec + tn sec + sec tn sec d= sec d = d sec + tn sec + tn = ln sec + tn +C u = sec + tn du = (sec + sec tn )d

92 54 Chpter 7 / Principles of Integrl Evlution The following bsic integrls occur frequentl nd re worth noting: tn d= tn + C (3) sec d= tn + C (4) Formul (4) is lred known to us, since the derivtive of tn is sec. Formul (3) cn be obtined b ppling reduction formul (9) with n = (verif) or, lterntivel, b using the identit + tn = sec to write tn d= (sec )d = tn + C The formuls tn 3 d= tn ln sec +C (5) sec 3 d= sec tn + ln sec + tn +C (6) cn be deduced from (), (), nd reduction formuls (9) nd () s follows: tn 3 d= tn tn d= tn ln sec +C sec 3 d= sec tn + sec d= sec tn + ln sec + tn +C INTEGRATING PRODUCTS OF TANGENTS AND SECANTS If m nd n re positive integers, then the integrl tn m sec n d cn be evluted b one of the three procedures stted in Tble 7.3., depending on whether m nd n re odd or even. Tble 7.3. integrting products of tngents nd secnts tn m sec n d procedure relevnt identities Split off fctor of sec. n even Appl the relevnt identit. sec = tn + Mke the substitution u = tn. Split off fctor of sec tn. m odd Appl the relevnt identit. tn = sec Mke the substitution u = sec. m even n odd Use the relevnt identities to reduce the integrnd to powers of sec lone. Then use the reduction formul for powers of sec. tn = sec

93 7.3 Integrting Trigonometric Functions 55 Emple 4 Evlute () tn sec 4 d (b) tn 3 sec 3 d (c) tn sec d Solution (). Since n = 4 is even, we will follow the first procedure in Tble 7.3.: tn sec 4 d= tn sec sec d = tn (tn + ) sec d = u (u + )du = 5 u5 + 3 u3 + C = 5 tn5 + 3 tn3 + C Solution (b). Since m = 3 is odd, we will follow the second procedure in Tble 7.3.: tn 3 sec 3 d= tn sec (sec tn )d = (sec ) sec (sec tn )d = (u )u du = 5 u5 3 u3 + C = 5 sec5 3 sec3 + C Solution (c). Since m = isevenndn = is odd, we will follow the third procedure in Tble 7.3.: tn sec d= (sec ) sec d = sec 3 d sec d See (6) nd () = sec tn + ln sec + tn ln sec + tn +C = sec tn ln sec + tn +C With the id of the identit + cot = csc the techniques in Tble 7.3. cn be dpted to evlute integrls of the form cot m csc n d It is lso possible to derive reduction formuls for powers of cot nd csc tht re nlogous to Formuls (9) nd (). AN ALTERNATIVE METHOD FOR INTEGRATING POWERS OF SINE, COSINE, TANGENT, AND SECANT The methods in Tbles 7.3. nd 7.3. cn sometimes be pplied if m = orn = to integrte positive integer powers of sine, cosine, tngent, nd secnt without reduction formuls. For emple, insted of using the reduction formul to integrte sin 3, we cn ppl the second procedure in Tble 7.3.: sin 3 d= (sin )sin d which grees with (). = ( cos )sin d = ( u )du u = cos du = sin d = 3 u3 u + C = 3 cos3 cos + C

94 56 Chpter 7 / Principles of Integrl Evlution MERCATOR S MAP OF THE WORLD The integrl of sec pls n importnt role in the design of nvigtionl mps for chrting nuticl nd eronuticl courses. Silors nd pilots usull chrt their courses long pths with constnt compss hedings; for emple, the course might be 3 northest or 35 southest. Ecept for courses tht re prllel to the equtor or run due north or south, course with constnt compss heding spirls round the Erth towrd one of the poles (s in the top prt of Figure 7.3.). In 569 the Flemish mthemticin nd geogrpher Gerhrd Krmer (5 594) (better known b the Ltin nme Merctor) devised world mp, clled the Merctor projection, in which spirls of constnt compss hedings pper s stright lines. This ws etremel importnt becuse it enbled silors to determine compss hedings between two points b connecting them with stright line on mp (s in the bottom prt of Figure 7.3.). If the Erth is ssumed to be sphere of rdius 4 mi, then the lines of ltitude t increments re equll spced bout 7 mi prt (wh?). However, in the Merctor projection, the lines of ltitude become wider prt towrd the poles, so tht two widel spced ltitude lines ner the poles m be ctull the sme distnce prt on the Erth s two closel spced ltitude lines ner the equtor. It cn be proved tht on Merctor mp in which the equtoril line hs length L, the verticl distnce D β on the mp between the equtor (ltitude ) nd the line of ltitude β is D β = L π βπ/8 sec d (7) Figure 7.3. A flight pth with constnt compss heding from New York Cit to Moscow follows spirl towrd the North Pole but is stright line segment on Merctor projection. QUICK CHECK EXERCISES 7.3 (See pge 58 for nswers.). Complete ech trigonometric identit with n epression involving cos. () sin = (b) cos = (c) cos sin =. Evlute the integrl. () sec d= (b) tn d= (c) sec d= (d) tn d= 3. Use the indicted substitution to rewrite the integrl in terms of u. Do not evlute the integrl. () sin cos d; u = sin (b) sin 3 cos d; u = cos (c) tn 3 sec d; u = tn (d) tn 3 sec d; u = sec EXERCISE SET Evlute the integrl.. cos 3 sin d. 3. sin 5θdθ 4. sin 5 3 cos 3d cos 3d sin 3 θ dθ 6. sin cos d 8. sin t cos 3 tdt. cos 3 t dt sin 3 cos 3 d sin 3 cos d

95 π/ π/3 π/6 sin cos d. sin cos 3d 4. sin cos(/)d 6. cos 3 d 8. sin 4 3 cos 3 3d. sin 4 cos d. sec ( )d 4. e tn(e )d 6. sec 4d 8. tn sec d 3. tn 4 sec 4 4d 3. sec 5 tn 3 d 34. tn 4 sec d 36. tn t sec 3 tdt 38. sec 4 d 4. tn 3 4d 4. tn sec 4 d 44. π/8 π/ tn d 46. π/ π π π sin cos 4 d sin 3θ cos θ dθ cos /3 sin d sin cos d cos 5θdθ sin k d tn 5d cot 3d sec( ) d tn 5 sec 4 d tn 4 θ sec 4 θdθ tn 5 θ sec θdθ tn sec 3 d tn sec 5 d sec 5 d tn 4 d tn sec 3/ d π/6 sec 3 θ tn θ dθ 47. tn 5 /4 d 48. sec π tn π d 49. cot 3 csc 3 d 5. cot 3t sec 3t dt 5. cot 3 d 5. csc 4 d True Flse Determine whether the sttement is true or flse. Eplin our nswer. 53. To evlute sin 5 cos 8 d, use the trigonometric identit sin = cos nd the substitution u = cos. 54. To evlute sin 8 cos 5 d, use the trigonometric identit sin = cos nd the substitution u = cos. 7.3 Integrting Trigonometric Functions The trigonometric identit sin α cos β = [sin(α β) + sin(α + β)] is often useful for evluting integrls of the form sin m cos n d. 56. The integrl tn 4 sec 5 d is equivlent to one whose integrnd is polnomil in sec. 57. Let m, n be distinct nonnegtive integers. Use Formuls (6) (8) to prove: () (b) (c) π π π sin m cos n d = cos m cos n d = sin m sin n d =. 58. Evlute the integrls in Eercise 57 when m nd n denote the sme nonnegtive integer. 59. Find the rc length of the curve = ln(cos ) over the intervl [,π/4]. 6. Find the volume of the solid generted when the region enclosed b = tn, =, nd = is revolved bout the -is. 6. Find the volume of the solid tht results when the region enclosed b = cos, = sin, =, nd = π/4 is revolved bout the -is. 6. The region bounded below b the -is nd bove b the portion of = sin from = to = π is revolved bout the -is. Find the volume of the resulting solid. 63. Use Formul (7) to show tht if the length of the equtoril line on Merctor projection is L, then the verticl distnce D between the ltitude lines t α nd β on the sme side of the equtor (where α<β)is D = L π ln sec β + tn β sec α + tn α 64. Suppose tht the equtor hs length of cm on Merctor projection. In ech prt, use the result in Eercise 63 to nswer the question. () Wht is the verticl distnce on the mp between the equtor nd the line t 5 north ltitude? (b) Wht is the verticl distnce on the mp between New Orlens, Louisin, t 3 north ltitude nd Winnipeg, Cnd, t 5 north ltitude? FOCUS ON CONCEPTS 65. () Show tht csc d= ln csc + cot +C (b) Show tht the result in prt () cn lso be written s csc d= ln csc cot +C nd csc d= ln tn + C

96 58 Chpter 7 / Principles of Integrl Evlution 66. Rewrite sin + cos in the form A sin( + φ) nd use our result together with Eercise 65 to evlute d sin + cos 67. Use the method of Eercise 66 to evlute d (, b not both zero) sin + b cos 68. () Use Formul (9) in Section 7. to show tht π/ π/ sin n d= n sin n d (n ) n (b) Use this result to derive the Wllis sine formuls: π/ sin n d= π ( ) 3 5 (n ) n even 4 6 n nd π/ ( ) n odd nd 3 sin n 4 6 (n ) d= n 69. Use the Wllis formuls in Eercise 68 to evlute () π/ sin 3 d (b) π/ sin 4 d (c) π/ sin 5 d (d) π/ sin 6 d. 7. Use Formul () in Section 7. nd the method of Eercise 68 to derive the Wllis cosine formuls: π/ cos n d= π ( ) 3 5 (n ) n even 4 6 n nd π/ cos n d= 4 6 (n ) n ( ) n odd nd 3 7. Writing Describe the vrious pproches for evluting integrls of the form sin m cos n d Into wht cses do these tpes of integrls fll? Wht procedures nd identities re used in ech cse? 7. Writing Describe the vrious pproches for evluting integrls of the form tn m sec n d Into wht cses do these tpes of integrls fll? Wht procedures nd identities re used in ech cse? QUICK CHECK ANSWERS 7.3. () 3. () cos + cos (b) (c) cos. () tn + C (b) tn + C (c) ln sec + tn +C (d) ln sec +C u du (b) (u )u du (c) u 3 du (d) (u )du 7.4 TRIGONOMETRIC SUBSTITUTIONS In this section we will discuss method for evluting integrls contining rdicls b mking substitutions involving trigonometric functions. We will lso show how integrls contining qudrtic polnomils cn sometimes be evluted b completing the squre. THE METHOD OF TRIGONOMETRIC SUBSTITUTION To strt, we will be concerned with integrls tht contin epressions of the form, +, in which is positive constnt. The bsic ide for evluting such integrls is to mke substitution for tht will eliminte the rdicl. For emple, to eliminte the rdicl in the epression, we cn mke the substitution = sin θ, π/ θ π/ () which ields = sin θ = ( sin θ) = cos θ = cos θ = cos θ cos θ since π/ θ π/

97 7.4 Trigonometric Substitutions 59 The restriction on θ in () serves two purposes it enbles us to replce cos θ b cos θ to simplif the clcultions, nd it lso ensures tht the substitutions cn be rewritten s θ = sin (/), if needed. Emple Evlute d 4. Solution. This ields To eliminte the rdicl we mke the substitution = sin θ, d = cos θdθ d = cos θdθ 4 ( sin θ) 4 4 sin θ cos θdθ = ( sin θ) ( cos θ) = dθ 4 sin θ = csc θdθ= cot θ + C () 4 4 u 4 = sin u Figure 7.4. At this point we hve completed the integrtion; however, becuse the originl integrl ws epressed in terms of, it is desirble to epress cot θ in terms of s well. This cn be done using trigonometric identities, but the epression cn lso be obtined b writing the substitution = sin θ s sin θ = / nd representing it geometricll s in Figure From tht figure we obtin 4 cot θ = Substituting this in () ields d 4 = C Emple d Evlute. 4 Solution. There re two possible pproches: we cn mke the substitution in the indefinite integrl (s in Emple ) nd then evlute the definite integrl using the -limits of integrtion, or we cn mke the substitution in the definite integrl nd convert the -limits to the corresponding θ-limits. Method. Using the result from Emple with the -limits of integrtion ields [ ] d 4 = 4 = [ ] 3 3 = Method. The substitution = sin θ cn be epressed s / = sin θ or θ = sin (/), so the θ-limits tht correspond to = nd = re = : θ = sin (/) = π/6 = : θ = sin ( /) = π/4

98 5 Chpter 7 / Principles of Integrl Evlution Thus, from () in Emple we obtin d 4 = 4 = 4 π/4 π/6 csc θdθ [ cot θ ] π/4 π/6 = 4 Convert -limits to θ-limits. [ ] 3 3 = 4 Emple 3 Find the re of the ellipse + b = b Solution. Becuse the ellipse is smmetric bout both es, its re A is four times the re in the first qudrnt (Figure 7.4.). If we solve the eqution of the ellipse for in terms of, we obtin =± b + b = Figure 7.4. = Figure where the positive squre root gives the eqution of the upper hlf. Thus, the re A is given b b A = 4 d = 4b d To evlute this integrl, we will mke the substitution = sin θ (so d = cos θdθ) nd convert the -limits of integrtion to θ-limits. Since the substitution cn be epressed s θ = sin (/), the θ-limits of integrtion re Thus, we obtin A = 4b = 4b = 4b π/ π/ = : θ = sin () = = : θ = sin () = π/ d = 4b cos θ cos θdθ π/ π/ cos θdθ= 4b sin θ cos θdθ ( + cos θ)dθ = b [θ + ] π/ [ π ] sin θ = b = πb TECHNOLOGY MASTERY REMARK If ou hve clculting utilit with numericl integrtion cpbilit, use it nd Formul (3) to pproimte π to three deciml plces. In the specil cse where = b, the ellipse becomes circle of rdius, nd the re formul becomes A = π, s epected. It is worth noting tht d = π (3) since this integrl represents the re of the upper semicircle (Figure 7.4.3). Thus fr, we hve focused on using the substitution = sin θ to evlute integrls involving rdicls of the form. Tble 7.4. summrizes this method nd describes some other substitutions of this tpe.

99 7.4 Trigonometric Substitutions 5 Tble 7.4. trigonometric substitutions epression in the integrnd substitution restriction on u simplifiction + = sin u c/ u c/ = tn u c/ < u < c/ = sin u = cos u + = + tn u = sec u = sec u u < c/ (if ) c/ < u c (if ) = sec u = tn u = / Find the rc length of the curve = / from = to = (Fig- Emple 4 ure 7.4.4). Solution. From Formul (4) of Section 6.4 the rc length L of the curve is ( ) d L = + d = + d d Figure The integrnd involves rdicl of the form + with =, so from Tble 7.4. we mke the substitution = tn θ, π/ <θ<π/ d dθ = sec θ or d = sec θdθ Since this substitution cn be epressed s θ = tn, the θ-limits of integrtion tht correspond to the -limits, = nd =, re Thus, L = + d = = = = = = = : θ = tn = = : θ = tn = π/4 π/4 π/4 π/4 π/4 + tn θ sec θdθ sec θ sec θdθ sec θ sec θdθ sec 3 θdθ + tn θ = sec θ sec θ> since π/ <θ<π/ [ ] π/4 sec θ tn θ + ln sec θ + tn θ [ ] + ln( + ).48 Formul (6) of Section 7.3 Emple 5 5 Evlute d, ssuming tht 5.

100 5 Chpter 7 / Principles of Integrl Evlution Solution. The integrnd involves rdicl of the form with = 5, so from Tble 7.4. we mke the substitution = 5 sec θ, θ<π/ d = 5 sec θ tn θ dθ or d = 5 sec θ tn θdθ Thus, 5 5 sec θ 5 d = (5 sec θ tn θ)dθ 5 sec θ 5 tn θ = (5 sec θ tn θ)dθ 5 sec θ = 5 tn θdθ tn θ since θ<π/ = 5 (sec θ )dθ = 5 tn θ 5θ + C u 5 = 5 sec u Figure To epress the solution in terms of, we will represent the substitution = 5 sec θ geometricll b the tringle in Figure 7.4.5, from which we obtin 5 tn θ = 5 From this nd the fct tht the substitution cn be epressed s θ = sec (/5), we obtin 5 d = ( 5 5 sec ) + C 5 INTEGRALS INVOLVING + b + c Integrls tht involve qudrtic epression + b + c, where = nd b =, cn often be evluted b first completing the squre, then mking n pproprite substitution. The following emple illustrtes this ide. Emple 6 Solution. Thus, the substitution ields Evlute d. Completing the squre ields = ( 4 + 4) = ( ) d = = = u =, du = d ( ) + 4 d = u + u + 4 du u u + 4 du + du u + 4 u u + 4 du + du u + 4 = ln(u + 4) + ( ) tn u + C = ln[( ) + 4]+tn ( ) + C

101 7.4 Trigonometric Substitutions 53 QUICK CHECK EXERCISES 7.4 (See pge 54 for nswers.). For ech epression, give trigonometric substitution tht will eliminte the rdicl. () (b) + (c). If = sec θ nd <θ<π/, then () sin θ = (b) cos θ = (c) tn θ =. 3. In ech prt, stte the trigonometric substitution tht ou would tr first to evlute the integrl. Do not evlute the integrl. 9 () + d 9 (b) d (c) 9 d (d) (e) (f ) 9 d d + (9) d 4. In ech prt, determine the substitution u. () + d = u + 3 du; u = u (b) d = du; u = 4 (c) 4 d = u du; u = EXERCISE SET 7.4 C CAS 6 Evlute the integrl.. 4 d. 4 d d 3. d d (4 + ) 5 d + 9 d 7. d d. 3 5 d d + t.. dt 9 4 t d d ( ) 3/ + 5 d d (4 9) 3/ 9. e e d d. d 4. / d d 5 cos θ dθ sin θ d ( ) 4 d 5. 3 d d (3 + ) 5/ 7 3 True Flse Determine whether the sttement is true or flse. Eplin our nswer. 7. An integrnd involving rdicl of the form suggests the substitution = sin θ. 8. The trigonometric substitution = sin θ is mde with the restriction θ π. 9. An integrnd involving rdicl of the form suggests the substitution = cos θ. 3. The re enclosed b the ellipse + 4 = isπ/. FOCUS ON CONCEPTS 3. The integrl + 4 d cn be evluted either b trigonometric substitution or b the substitution u = + 4. Do it both ws nd show tht the results re equivlent. 3. The integrl + 4 d cn be evluted either b trigonometric substitution or b lgebricll rewriting the numertor of the integrnd s ( + 4) 4. Do it both ws nd show tht the results re equivlent. 33. Find the rc length of the curve = ln from = to =. 34. Find the rc length of the curve = from = to =.

102 C 54 Chpter 7 / Principles of Integrl Evlution 35. Find the re of the surfce generted when the curve in Eercise 34 is revolved bout the -is. 36. Find the volume of the solid generted when the region enclosed b = ( ) /4, =, =, nd = is revolved bout the -is Evlute the integrl. d d d d 44. d d d d d e d + e + e d (4 )d 49 5 There is good chnce tht our CAS will not be ble to evlute these integrls s stted. If this is so, mke substitution tht converts the integrl into one tht our CAS cn evlute. 49. cos sin sin 4 d 5. ( cos + sin ) + sin d 5. () Use the hperbolic substitution = 3 sinh u, the identit cosh u sinh u =, nd Theorem to evlute d + 9 (b) Evlute the integrl in prt () using trigonometric substitution nd show tht the result grees with tht obtined in prt (). 5. Use the hperbolic substitutionn = cosh u, the identit sinh u = (cosh u ), nd the results referenced in Eercise 5 to evlute d, 53. Writing The trigonometric substitution = sin θ, π/ θ π/, is suggested for n integrl whose integrnd involves. Discuss the implictions of restricting θ to π/ θ 3π/, nd eplin wh the restriction π/ θ π/ should be preferred. 54. Writing The trigonometric substitution = cos θ could lso be used for n integrl whose integrnd involves. Determine n pproprite restriction for θ with the substitution = cos θ, nd discuss how to ppl this substitution in pproprite integrls. Illustrte our discussion b evluting the integrl in Emple using substitution of this tpe. QUICK CHECK ANSWERS () = sin θ (b) = tn θ (c) = sec θ. () (b) (c) 4 3. () = 3 tn θ (b) = 3 sin θ (c) = 3 sin θ (d) = 3 sec θ (e) = 3 tn θ (f ) = tn θ 4. () (b) 3 (c) INTEGRATING RATIONAL FUNCTIONS BY PARTIAL FRACTIONS Recll tht rtionl function is rtio of two polnomils. In this section we will give generl method for integrting rtionl functions tht is bsed on the ide of decomposing rtionl function into sum of simple rtionl functions tht cn be integrted b the methods studied in erlier sections. PARTIAL FRACTIONS In lgebr, one lerns to combine two or more frctions into single frction b finding common denomintor. For emple, ( + ) + 3( 4) 5 = = + ( 4)( + ) 3 4 ()

103 7.5 Integrting Rtionl Functions b Prtil Frctions 55 However, for purposes of integrtion, the left side of () is preferble to the right side since ech of the terms is es to integrte: d = 4 d + 3 d = ln 4 +3ln + +C + Thus, it is desirble to hve some method tht will enble us to obtin the left side of (), strting with the right side. To illustrte how this cn be done, we begin b noting tht on the left side the numertors re constnts nd the denomintors re the fctors of the denomintor on the right side. Thus, to find the left side of (), strting from the right side, we could fctor the denomintor of the right side nd look for constnts A nd B such tht 5 ( 4)( + ) = A 4 + B () + One w to find the constnts A nd B is to multipl () through b ( 4)( + ) to cler frctions. This ields 5 = A( + ) + B( 4) (3) This reltionship holds for ll, so it holds in prticulr if = 4or =. Substituting = 4 in (3) mkes the second term on the right drop out nd ields the eqution = 5A or A = ; nd substituting = in (3) mkes the first term on the right drop out nd ields the eqution 5 = 5B or B = 3. Substituting these vlues in () we obtin 5 ( 4)( + ) = (4) + which grees with (). A second method for finding the constnts A nd B is to multipl out the right side of (3) nd collect like powers of to obtin 5 = (A + B) + (A 4B) Since the polnomils on the two sides re identicl, their corresponding coefficients must be the sme. Equting the corresponding coefficients on the two sides ields the following sstem of equtions in the unknowns A nd B: A + B = 5 A 4B = Solving this sstem ields A = nd B = 3 s before (verif). The terms on the right side of (4) re clled prtil frctions of the epression on the left side becuse the ech constitute prt of tht epression. To find those prtil frctions we first hd to mke guess bout their form, nd then we hd to find the unknown constnts. Our net objective is to etend this ide to generl rtionl functions. For this purpose, suppose tht P()/Q() is proper rtionl function, b which we men tht the degree of the numertor is less thn the degree of the denomintor. There is theorem in dvnced lgebr which sttes tht ever proper rtionl function cn be epressed s sum P() Q() = F () + F () + +F n () where F (), F (),...,F n () re rtionl functions of the form A ( + b) k or A + B ( + b + c) k in which the denomintors re fctors of Q(). The sum is clled the prtil frction decomposition of P()/Q(), nd the terms re clled prtil frctions. As in our opening emple, there re two prts to finding prtil frction decomposition: determining the ect form of the decomposition nd finding the unknown constnts.

104 56 Chpter 7 / Principles of Integrl Evlution FINDING THE FORM OF A PARTIAL FRACTION DECOMPOSITION The first step in finding the form of the prtil frction decomposition of proper rtionl function P()/Q() is to fctor Q() completel into liner nd irreducible qudrtic fctors, nd then collect ll repeted fctors so tht Q() is epressed s product of distinct fctors of the form ( + b) m nd ( + b + c) m From these fctors we cn determine the form of the prtil frction decomposition using two rules tht we will now discuss. LINEAR FACTORS If ll of the fctors of Q() re liner, then the prtil frction decomposition of P()/Q() cn be determined b using the following rule: liner fctor rule For ech fctor of the form ( + b) m, the prtil frction decomposition contins the following sum of m prtil frctions: A + b + A ( + b) + + A m ( + b) m where A, A,...,A m re constnts to be determined. In the cse where m =, onl the first term in the sum ppers. Emple Evlute d +. Solution. The integrnd is proper rtionl function tht cn be written s + = ( )( + ) The fctors nd + re both liner nd pper to the first power, so ech contributes one term to the prtil frction decomposition b the liner fctor rule. Thus, the decomposition hs the form ( )( + ) = A + B + where A nd B re constnts to be determined. Multipling this epression through b ( )( + ) ields = A( + ) + B( ) (6) As discussed erlier, there re two methods for finding A nd B: we cn substitute vlues of tht re chosen to mke terms on the right drop out, or we cn multipl out on the right nd equte corresponding coefficients on the two sides to obtin sstem of equtions tht cn be solved for A nd B. We will use the first pproch. Setting = mkes the second term in (6) drop out nd ields = 3A or A = 3 ; nd setting = mkes the first term in (6) drop out nd ields = 3B or B = 3. Substituting these vlues in (5) ields the prtil frction decomposition ( )( + ) = (5)

105 7.5 Integrting Rtionl Functions b Prtil Frctions 57 The integrtion cn now be completed s follows: d ( )( + ) = d 3 d 3 + = 3 ln 3 ln + +C = 3 ln + + C If the fctors of Q() re liner nd none re repeted, s in the lst emple, then the recommended method for finding the constnts in the prtil frction decomposition is to substitute pproprite vlues of to mke terms drop out. However, if some of the liner fctors re repeted, then it will not be possible to find ll of the constnts in this w. In this cse the recommended procedure is to find s mn constnts s possible b substitution nd then find the rest b equting coefficients. This is illustrted in the net emple. Emple Evlute d. Solution. The integrnd cn be rewritten s = + 4 ( ) Although is qudrtic fctor, it is not irreducible since =. Thus, b the liner fctor rule, introduces two terms (since m = ) of the form A + B nd the fctor introduces one term (since m = ) of the form C so the prtil frction decomposition is + 4 ( ) = A + B + C (7) Multipling b ( ) ields + 4 = A( ) + B( ) + C (8) which, fter multipling out nd collecting like powers of, becomes + 4 = (A + C) + ( A + B) B (9) Setting = in (8) mkes the first nd third terms drop out nd ields B =, nd setting = in (8) mkes the first nd second terms drop out nd ields C = (verif). However, there is no substitution in (8) tht produces A directl, so we look to Eqution (9) to find this vlue. This cn be done b equting the coefficients of on the two sides to obtin A + C = or A = C = Substituting the vlues A =, B =, nd C = in (7) ields the prtil frction decomposition + 4 ( ) = + + Thus, + 4 d d d = ( ) + d = ln + + ln +C = ln + + C

106 58 Chpter 7 / Principles of Integrl Evlution QUADRATIC FACTORS If some of the fctors of Q() re irreducible qudrtics, then the contribution of those fctors to the prtil frction decomposition of P()/Q() cn be determined from the following rule: qudrtic fctor rule For ech fctor of the form ( + b + c) m, the prtil frction decomposition contins the following sum of m prtil frctions: A + B + b + c + A + B ( + b + c) + + A m + B m ( + b + c) m where A, A,...,A m, B, B,...,B m re constnts to be determined. In the cse where m =, onl the first term in the sum ppers. Emple 3 Evlute d. Solution. The denomintor in the integrnd cn be fctored b grouping: = (3 ) + (3 ) = (3 )( + ) B the liner fctor rule, the fctor 3 introduces one term, nmel, A 3 nd b the qudrtic fctor rule, the fctor + introduces one term, nmel, B + C + Thus, the prtil frction decomposition is Multipling b (3 )( + ) ields + (3 )( + ) = A 3 + B + C + () + = A( + ) + (B + C)(3 ) () We could find A b substituting = to mke the lst term drop out, nd then find the rest 3 of the constnts b equting corresponding coefficients. However, in this cse it is just s es to find ll of the constnts b equting coefficients nd solving the resulting sstem. For this purpose we multipl out the right side of () nd collect like terms: + = (A + 3B) + ( B + 3C) + (A C) Equting corresponding coefficients gives A + 3B = B + 3C = A C = To solve this sstem, subtrct the third eqution from the first to eliminte A. Then use the resulting eqution together with the second eqution to solve for B nd C. Finll, determine A from the first or third eqution. This ields (verif) A = 7 5, B = 4 5, C = 3 5

107 7.5 Integrting Rtionl Functions b Prtil Frctions 59 TECHNOLOGY MASTERY Computer lgebr sstems hve builtin cpbilities for finding prtil frction decompositions. If ou hve CAS, use it to find the decompositions in Emples,, nd 3. Thus, () becomes nd + (3 )( + ) d = (3 )( + ) = d d d + = 7 5 ln ln( + ) tn + C Emple Evlute d. ( + )( + 3) Solution. Observe tht the integrnd is proper rtionl function since the numertor hs degree 4 nd the denomintor hs degree 5. Thus, the method of prtil frctions is pplicble. B the liner fctor rule, the fctor + introduces the single term A + nd b the qudrtic fctor rule, the fctor ( + 3) introduces two terms (since m = ): B + C D + E ( + 3) Thus, the prtil frction decomposition of the integrnd is ( + )( + 3) = A + + B + C D + E ( + 3) () Multipling b ( + )( + 3) ields = A( + 3) + (B + C)( + 3)( + ) + (D + E)( + ) (3) which, fter multipling out nd collecting like powers of, becomes = (A + B) 4 + (B + C) 3 + (6A + 3B + C + D) + (6B + 3C + D + E) + (9A + 6C + E) (4) Equting corresponding coefficients in (4) ields the following sstem of five liner equtions in five unknowns: A + B = 3 B + C = 4 6A + 3B + C + D = 6 (5) 6B + 3C + D + E = 9A + 6C + E = 9 Efficient methods for solving sstems of liner equtions such s this re studied in brnch of mthemtics clled liner lgebr; those methods re outside the scope of this tet. However, s prcticl mtter most liner sstems of n size re solved b computer, nd most computer lgebr sstems hve commnds tht in mn cses cn solve liner sstems ectl. In this prticulr cse we cn simplif the work b first substituting =

108 5 Chpter 7 / Principles of Integrl Evlution in (3), which ields A =. Substituting this known vlue of A in (5) ields the simpler sstem B = B + C = 4 3B + C + D = 6B + 3C + D + E = 6C + E = This sstem cn be solved b strting t the top nd working down, first substituting B = in the second eqution to get C =, then substituting the known vlues of B nd C in the third eqution to get D = 4, nd so forth. This ields Thus, () becomes nd so A =, B =, C =, D = 4, E = ( + )( + 3) = d ( + )( + 3) d = d = ln + +ln( + 3) C 4 ( + 3) ( + 3) d INTEGRATING IMPROPER RATIONAL FUNCTIONS Although the method of prtil frctions onl pplies to proper rtionl functions, n improper rtionl function cn be integrted b performing long division nd epressing the function s the quotient plus the reminder over the divisor. The reminder over the divisor will be proper rtionl function, which cn then be decomposed into prtil frctions. This ide is illustrted in the following emple. (6) Emple Evlute d. + Solution. The integrnd is n improper rtionl function since the numertor hs degree 4 nd the denomintor hs degree. Thus, we first perform the long division It follows tht the integrnd cn be epressed s = (3 + ) + + nd hence d = + (3 + )d+ d +

109 7.5 Integrting Rtionl Functions b Prtil Frctions 5 The second integrl on the right now involves proper rtionl function nd cn thus be evluted b prtil frction decomposition. Using the result of Emple we obtin d = ln + + C CONCLUDING REMARKS There re some cses in which the method of prtil frctions is inpproprite. For emple, it would be inefficient to use prtil frctions to perform the integrtion d = ln C since the substitution u = is more direct. Similrl, the integrtion + d = + d d + = ln( + ) tn + C requires onl little lgebr since the integrnd is lred in prtil frction form. QUICK CHECK EXERCISES 7.5 (See pge 53 for nswers.). A prtil frction is rtionl function of the form or of the form.. () Wht is proper rtionl function? (b) Wht condition must the degree of the numertor nd the degree of the denomintor of rtionl function stisf for the method of prtil frctions to be pplicble directl? (c) If the condition in prt (b) is not stisfied, wht must ou do if ou wnt to use prtil frctions? 3. Suppose tht the function f() = P()/Q() is proper rtionl function. () For ech fctor of Q() of the form ( + b) m, the prtil frction decomposition of f contins the following sum of m prtil frctions: (b) For ech fctor of Q() of the form ( + b + c) m, where + b + c is n irreducible qudrtic, the prtil frction decomposition of f contins the following sum of m prtil frctions: 4. Complete the prtil frction decomposition. 3 () ( + )( ) = A + 3 (b) ( + )(3 + ) = B Evlute the integrl. 3 3 () d (b) ( + )( ) ( + )(3 + ) d EXERCISE SET 7.5 C CAS 8 Write out the form of the prtil frction decomposition. (Do not find the numericl vlues of the coefficients.) ( 3)( + 4) ( + ) ( + 5) Evlute the integrl. d ( 4) ( + ) 3 3 ( )( + 6) 3 4 ( )( + ) d d d 9 9 d d ( ) 8 + d d 3 d d d d d. d d 4. d ( ) 3

110 5 Chpter 7 / Principles of Integrl Evlution C d 6. d ( + )( 3) d 8. d ( + ) 3 ( + ) 3 d 9. d 3. (4 )( + ) d 3. ( + )( + 3) ( + )( + ) d d d True Flse Determine whether the sttement is true or flse. Eplin our nswer. 35. The technique of prtil frctions is used for integrls whose integrnds re rtios of polnomils. 36. The integrnd in ( + ) d is proper rtionl function. 37. The prtil frction decomposition of + 3 is If f() = P()/( + 5) 3 is proper rtionl function, then the prtil frction decomposition of f() hs terms with constnt numertors nd denomintors ( + 5), ( + 5), nd ( + 5) Evlute the integrl b mking substitution tht converts the integrnd to rtionl function. cos θ e t 39. sin dθ 4. θ + 4 sin θ 5 e t 4 dt e ln 4. d 4. e + 4 ( + ln ) d 43. Find the volume of the solid generted when the region enclosed b = /(9 ), =, =, nd = is revolved bout the -is. 44. Find the re of the region under the curve = /( + e ), over the intervl [ ln 5, ln 5]. [Hint: Mke substitution tht converts the integrnd to rtionl function.] Use CAS to evlute the integrl in two ws: (i) integrte directl; (ii) use the CAS to find the prtil frction decomposition nd integrte the decomposition. Integrte b hnd to check the results ( + + 3) d d ( + ) 3 C Integrte b hnd nd check our nswers using CAS. d d FOCUS ON CONCEPTS 49. Show tht 4 + d = π 8 5. Use prtil frctions to derive the integrtion formul d = ln + + C 5. Suppose tht + b + c is qudrtic polnomil nd tht the integrtion + b + c d produces function with no inverse tngent terms. Wht does this tell ou bout the roots of the polnomil? 5. Suppose tht + b + c is qudrtic polnomil nd tht the integrtion + b + c d produces function with neither logrithmic nor inverse tngent terms. Wht does this tell ou bout the roots of the polnomil? 53. Does there eist qudrtic polnomil + b + c such tht the integrtion + b + c d produces function with no logrithmic terms? If so, give n emple; if not, eplin wh no such polnomil cn eist. 54. Writing Suppose tht P() is cubic polnomil. Stte the generl form of the prtil frction decomposition for f() = P() ( + 5) 4 nd stte the implictions of this decomposition for evluting the integrl f()d. 55. Writing Consider the functions f() = nd g() = 4 4 Ech of the integrls f()d nd g() d cn be evluted using prtil frctions nd using t lest one other integrtion technique. Demonstrte two different techniques for evluting ech of these integrls, nd then discuss the considertions tht would determine which technique ou would use.

111 7.6 Using Computer Algebr Sstems nd Tbles of Integrls 53 QUICK CHECK ANSWERS 7.5 A. ( + b) ; A + B. () A proper rtionl function is rtionl function in which the degree of the numertor is k ( + b + c) k less thn the degree of the denomintor. (b) The degree of the numertor must be less thn the degree of the denomintor. (c) Divide the denomintor into the numertor, which results in the sum of polnomil nd proper rtionl function. A 3. () + b + A ( + b) + + A m ( + b) (b) A + B m + b + c + A + B ( + b + c) + + A m + B m ( + b + c) m 4. () A = (b) B = 3 5. () d = ln + ( + )( ) + C (b) 3 ( + )(3 + ) d = 3 ln 3 + tn + C 7.6 USING COMPUTER ALGEBRA SYSTEMS AND TABLES OF INTEGRALS In this section we will discuss how to integrte using tbles, nd we will see some specil substitutions to tr when n integrl doesn t mtch n of the forms in n integrl tble. In prticulr, we will discuss method for integrting rtionl functions of sin nd cos. We will lso ddress some of the issues tht relte to using computer lgebr sstems for integrtion. Reders who re not using computer lgebr sstems cn skip tht mteril. INTEGRAL TABLES Tbles of integrls re useful for eliminting tedious hnd computtion. The endppers of this tet contin reltivel brief tble of integrls tht we will refer to s the Endpper Integrl Tble; more comprehensive tbles re published in stndrd reference books such s the CRC Stndrd Mthemticl Tbles nd Formule, CRC Press, Inc.,. All integrl tbles hve their own scheme for clssifing integrls ccording to the form of the integrnd. For emple, the Endpper Integrl Tble clssifies the integrls into 5 ctegories; Bsic Functions, Reciprocls of Bsic Functions, Powers of Trigonometric Functions, Products of Trigonometric Functions, nd so forth. The first step in working with tbles is to red through the clssifictions so tht ou understnd the clssifiction scheme nd know where to look in the tble for integrls of different tpes. PERFECT MATCHES If ou re luck, the integrl ou re ttempting to evlute will mtch up perfectl with one of the forms in the tble. However, when looking for mtches ou m hve to mke n djustment for the vrible of integrtion. For emple, the integrl sin d is perfect mtch with Formul (46) in the Endpper Integrl Tble, ecept for the letter used for the vrible of integrtion. Thus, to ppl Formul (46) to the given integrl we need to chnge the vrible of integrtion in the formul from u to. With tht minor modifiction we obtin sin d= sin + ( ) cos + C Here re some more emples of perfect mtches.

112 54 Chpter 7 / Principles of Integrl Evlution Emple Use the Endpper Integrl Tble to evlute () sin 7 cos d (b) 7 + 3d (c) d (d) ( ) sin π d Solution (). The integrnd cn be clssified s product of trigonometric functions. Thus, from Formul (4) with m = 7 nd n = we obtin cos 9 cos 5 sin 7 cos d= + C 8 Solution (b). The integrnd cn be clssified s power of multipling + b. Thus, from Formul (3) with = 7 nd b = 3 we obtin 7 + 3d= 835 ( )(7 + 3) 3/ + C Solution (c). The integrnd cn be clssified s power of dividing. Thus, from Formul (79) with = we obtin d = + ln + C Solution (d). The integrnd cn be clssified s polnomil multipling trigonometric function. Thus, we ppl Formul (58) with p() = nd = π. The successive nonzero derivtives of p() re nd so ( ) sin π d = π p () = 3 + 7, p () = 6, p () = 6 cos π π sin π + 6 π cos π 6 sin π + C 3 π4 MATCHES REQUIRING SUBSTITUTIONS Sometimes n integrl tht does not mtch n tble entr cn be mde to mtch b mking n pproprite substitution. Emple Use the Endpper Integrl Tble to evlute () e π sin (e π )d (b) d Solution (). The integrnd does not even come close to mtching n of the forms in the tble. However, little thought suggests the substitution u = e π, du = πe π d

113 7.6 Using Computer Algebr Sstems nd Tbles of Integrls 55 from which we obtin e π sin (e π )d = π sin udu The integrnd is now bsic function, nd Formul (7) ields e π sin (e π )d = π = π [u sin u + u ] + C [e π sin (e π ) + e π ] + C Solution (b). Agin, the integrnd does not closel mtch n of the forms in the tble. However, little thought suggests tht it m be possible to bring the integrnd closer to the form + b completing the squre to eliminte the term involving inside the rdicl. Doing this ields d = ( 4 + 4) + d = ( ) + d () At this point we re closer to the form +, but we re not quite there becuse of the ( ) rther thn inside the rdicl. However, we cn resolve tht problem with the substitution u =, du = d With this substitution we hve = u +, so () cn be epressed in terms of u s d = (u + ) u + du = u u u + du + + du The first integrl on the right is now perfect mtch with Formul (84) with =, nd the second is perfect mtch with Formul (7) with =. Thus, ppling these formuls we obtin d = 3 (u + ) 3/ + [ u u + + ln(u + ] u + ) + C If we now replce u b (in which cse u + = 4 + 5), we obtin d = 3 ( 4 + 5) 3/ + ( ) ( + ln + ) C Although correct, this form of the nswer hs n unnecessr miture of rdicls nd frctionl eponents. If desired, we cn clen up the nswer b writing ( 4 + 5) 3/ = ( 4 + 5) from which it follows tht (verif) d = 3 ( ) ln ( + ) C MATCHES REQUIRING REDUCTION FORMULAS In cses where the entr in n integrl tble is reduction formul, tht formul will hve to be pplied first to reduce the given integrl to form in which it cn be evluted.

114 56 Chpter 7 / Principles of Integrl Evlution Emple 3 Use the Endpper Integrl Tble to evlute 3 + d. Solution. The integrnd cn be clssified s power of multipling the reciprocl of + b. Thus, from Formul (7) with =,b =, nd n = 3, followed b Formul (6), we obtin 3 d = d = [ ( ) ] + + C ( 3 = ) + + C 35 SPECIAL SUBSTITUTIONS The Endpper Integrl Tble hs numerous entries involving n eponent of 3/ or involving squre roots (eponent /), but it hs no entries with other frctionl eponents. However, integrls involving frctionl powers of cn often be simplified b mking the substitution u = /n in which n is the lest common multiple of the denomintors of the eponents. The resulting integrl will then involve integer powers of u. Emple 4 Evlute () + 3 d (b) + e d Solution (). The integrnd contins / nd /3, so we mke the substitution u = /6, from which we obtin = u 6, d = 6u 5 du Thus, B long division + 3 d = (u 6 ) / + (u 6 ) /3 (6u5 )du= 6 u 8 + u = u6 u 4 + u + + u from which it follows tht ( + 3 d = 6 u 6 u 4 + u + ) du + u = 6 7 u7 6 5 u5 + u 3 6u + 6 tn u + C u 8 + u du = 6 7 7/ /6 + / 6 /6 + 6 tn ( /6 ) + C Solution (b). The integrl does not mtch n of the forms in the Endpper Integrl Tble. However, the tble does include severl integrls contining + bu. This suggests the substitution u = e, from which we obtin = ln u, d = u du

115 7.6 Using Computer Algebr Sstems nd Tbles of Integrls 57 Tr finding the ntiderivtive in Emple 4(b) using the substitution u = + e Thus, from Formul () with = nd b =, followed b Formul (8), we obtin + u + e d = du u = du + u + u + u = + u + u + ln + C + u + = [ ] + e + e + ln + C Absolute vlue not needed + e + + u / Figure 7.6. u Functions tht consist of finitel mn sums, differences, quotients, nd products of sin nd cos re clled rtionl functions of sin nd cos. Some emples re sin + 3 cos cos + 4 sin, sin + cos cos, 3 sin sin The Endpper Integrl Tble gives few formuls for integrting rtionl functions of sin nd cos under the heding Reciprocls of Bsic Functions. For emple, it follows from Formul (8) tht d = tn sec + C () + sin However, since the integrnd is rtionl function of sin, it m be desirble in prticulr ppliction to epress the vlue of the integrl in terms of sin nd cos nd rewrite () s sin d = + C + sin cos Mn rtionl functions of sin nd cos cn be evluted b n ingenious method tht ws discovered b the mthemticin Krl Weierstrss (see p. for biogrph). The ide is to mke the substitution from which it follows tht u = tn(/), π/ </ <π/ = tn u, d = + u du To implement this substitution we need to epress sin nd cos in terms of u. For this purpose we will use the identities sin = sin(/) cos(/) (3) cos = cos (/) sin (/) (4) nd the following reltionships suggested b Figure 7.6.: u sin(/) = nd cos(/) = + u + u Substituting these epressions in (3) nd (4) ields ( )( ) u sin = = u + u + u + u ( ) ( ) u cos = = u + u + u + u

116 58 Chpter 7 / Principles of Integrl Evlution In summr, we hve shown tht the substitution u = tn(/) cn be implemented in rtionl function of sin nd cos b letting sin = u u, cos = + u + u, d = du (5) + u Emple 5 Evlute d sin + cos. The substitution u = tn(/) will convert n rtionl function of sin nd cos to n ordinr rtionl function of u. However, the method cn led to cumbersome prtil frction decompositions, so it m be worthwhile to consider other methods s well when hnd computtions re being used. Solution. The integrnd is rtionl function of sin nd cos tht does not mtch n of the formuls in the Endpper Integrl Tble, so we mke the substitution u = tn(/). Thus, from (5) we obtin du d sin + cos = + u ( ) ( u u ) + + u + u du = ( + u ) u + ( u ) du = u = ln u +C = ln tn( /) +C INTEGRATING WITH COMPUTER ALGEBRA SYSTEMS Integrtion tbles re rpidl giving w to computerized integrtion using computer lgebr sstems. However, s with mn powerful tools, knowledgeble opertor is n importnt component of the sstem. Sometimes computer lgebr sstems do not produce the most generl form of the indefinite integrl. For emple, the integrl formul d = ln +C which cn be obtined b inspection or b using the substitution u =, is vlid for >orfor<. However, not ll computer lgebr sstems produce this form of the nswer. Some tpicl nswers produced b vrious implementtions of Mthemtic, Mple, nd the CAS on hndheld clcultor re ln( + ), ln( ), ln( ) Observe tht none of the sstems include the constnt of integrtion the nswer produced is prticulr ntiderivtive nd not the most generl ntiderivtive (indefinite integrl). Observe lso tht onl one of these nswers includes the bsolute vlue signs; the ntiderivtives produced b the other sstems re vlid onl for >. All sstems, however, re ble to clculte the definite integrl / d = ln correctl. Now let us emine how these sstems hndle the integrl d = 3 ( ) ln( ) (6)

117 7.6 Using Computer Algebr Sstems nd Tbles of Integrls 59 which we obtined in Emple (b) (with the constnt of integrtion included). Some CAS implementtions produce this result in slightl different lgebric forms, but version of Mple produces the result d = 3 ( 4 + 5) 3/ + ( 4) sinh ( ) This cn be rewritten s (6) b epressing the frctionl eponent in rdicl form nd epressing sinh ( ) in logrithmic form using Theorem (verif). A version of Mthemtic produces the result d = 3 ( ) sinh ( ) Epnding the epression ( + ) 8 8 produces constnt term of 8, wheres the second epression in (7) hs no constnt term. Wht is the eplntion? which cn be rewritten in form (6) b using Theorem together with the identit sinh ( ) = sinh (verif). Computer lgebr sstems cn sometimes produce inconvenient or unnturl nswers to integrtion problems. For emple, vrious computer lgebr sstems produced the following results when sked to integrte ( + ) 7 : ( + ) 8, (7) The first form is in keeping with the hnd computtion ( + ) 7 ( + )8 d = + C 8 tht uses the substitution u = +, wheres the second form is bsed on epnding ( + ) 7 nd integrting term b term. In Emple () of Section 7.3 we showed tht sin 4 cos 5 d= 5 sin5 7 sin7 + 9 sin9 + C However, version of Mthemtic integrtes this s 3 8 sin sin 3 sin 5 + sin 7 + sin wheres other computer lgebr sstems essentill integrte it s 9 sin3 cos 6 sin cos6 + 5 cos4 sin cos sin sin Although these three results look quite different, the cn be obtined from one nother using pproprite trigonometric identities. TECHNOLOGY MASTERY Sometimes integrls tht cnnot be evluted b CAS in their given form cn be evluted b first rewriting them in different form or b mking substitution. If ou hve CAS, mke u-substitution in (8) tht will enble ou to evlute the integrl with our CAS. Then evlute the integrl. COMPUTER ALGEBRA SYSTEMS HAVE LIMITATIONS A computer lgebr sstem combines set of integrtion rules (such s substitution) with librr of functions tht it cn use to construct ntiderivtives. Such librries contin elementr functions, such s polnomils, rtionl functions, trigonometric functions, s well s vrious nonelementr functions tht rise in engineering, phsics, nd other pplied fields. Just s our Endpper Integrl Tble hs onl indefinite integrls, these librries re not ehustive of ll possible integrnds. If the sstem cnnot mnipulte the integrnd to form mtching one in its librr, the progrm will give some indiction tht it cnnot evlute the integrl. For emple, when sked to evlute the integrl ( + ln ) + ( ln ) d (8) ll of the sstems mentioned bove respond b displing some form of the unevluted integrl s n nswer, indicting tht the could not perform the integrtion. Sometimes computer lgebr sstems respond b epressing n integrl in terms of nother integrl. For emple, if ou tr to integrte e using Mthemtic, Mple, or

118 53 Chpter 7 / Principles of Integrl Evlution Sge, ou will obtin n epression involving erf (which stnds for error function). The function erf() is defined s erf() = π e t dt so ll three progrms essentill rewrite the given integrl in terms of closel relted integrl. From one point of view this is wht we did in integrting /, since the nturl logrithm function is (formll) defined s (see Section 5.). ln = t dt Emple 6 time t is A prticle moves long n -is in such w tht its velocit v(t) t v(t) = 3 cos 7 t sin 4 t (t ) Grph the position versus time curve for the prticle, given tht the prticle is t = when t =. Solution. b Since d/dt = v(t) nd = when t =, the position function (t) is given (t) = + t v(s) ds Some computer lgebr sstems will llow this epression to be entered directl into commnd for plotting functions, but it is often more efficient to perform the integrtion first. The uthors integrtion utilit ields = 3 cos 7 t sin 4 tdt Figure 7.6. t = 3 sin t + sin 9 t 9 7 sin7 t + 6 sin 5 t + C where we hve dded the required constnt of integrtion. Using the initil condition () =, we substitute the vlues = nd t = into this eqution to find tht C =, so (t) = 3 sin t + sin 9 t 9 7 sin7 t + 6 sin 5 t + (t ) The grph of versus t is shown in Figure QUICK CHECK EXERCISES 7.6 (See pge 533 for nswers.). Find n integrl formul in the Endpper Integrl Tble tht cn be used to evlute the integrl. Do not evlute the integrl. () d (b) 5 4 d (c) 3 + d (d) ln d. In ech prt, mke the indicted u-substitution, nd then find n integrl formul in the Endpper Integrl Tble tht cn be used to evlute the integrl. Do not evlute the integrl. () d; u = + e (b) e d; u = e (c) d; u = + sin(e e ) (d) d; u = ( 4 ) 3/ 3. In ech prt, use the Endpper Integrl Tble to evlute the integrl. (If necessr, first mke n pproprite substitution or complete the squre.) (cont.)

119 7.6 Using Computer Algebr Sstems nd Tbles of Integrls 53 () (b) 4 d = cos cos d= (c) (d) e 3 e d = d = C C EXERCISE SET 7.6 C CAS 4 () Use the Endpper Integrl Tble to evlute the given integrl. (b) If ou hve CAS, use it to evlute the integrl, nd then confirm tht the result is equivlent to the one tht ou found in prt (). 4. d. 3 (4 5) d 3. d 4. ( + 5) ( 5) d d 6. d 7. d d 9. d. 6 9 d 5. 3 d. d 3. d 4. d d 6. d 4 7. d 8. 6 d 9. sin 3 sin 4d. sin cos 5d ln. 3 ln d. d 3 3. e sin 3d 4. e cos d 5 36 () Mke the indicted u-substitution, nd then use the Endpper Integrl Tble to evlute the integrl. (b) If ou hve CAS, use it to evlute the integrl, nd then confirm tht the result is equivlent to the one tht ou found in prt () e 4 d, u = (4 3e e ) sin d, u = cos (cos )(3 cos ) d, u = 3 (9 + 4) cos sin d, u = sin 4 4 C d, u = d, u = 4 5 d, u = 4 4 d, u = 3 4 sin (ln ) d, u = ln e cos (e )d, u= e e d, u = ln(3 + )d, u= () Mke n pproprite u-substitution, nd then use the Endpper Integrl Tble to evlute the integrl. (b) If ou hve CAS, use it to evlute the integrl (no substitution), nd then confirm tht the result is equivlent to tht in prt (). cos (sin 3)(sin 3 + ) d ln 38. 4ln d e 39. d 4. d e 4. e e d 4. d d 44. d sin d 46. cos d 47. e d 48. ln( + )d C 49 5 () Complete the squre, mke n pproprite u- substitution, nd then use the Endpper Integrl Tble to evlute the integrl. (b) If ou hve CAS, use it to evlute the integrl (no substitution or squre completion), nd then confirm tht the result is equivlent to tht in prt () d 5. d + 6 7

120 C C 53 Chpter 7 / Principles of Integrl Evlution 5. d d () Mke n pproprite u-substitution of the form u = /n or u = ( + ) /n, nd then evlute the integrl. (b) If ou hve CAS, use it to evlute the integrl, nd then confirm tht the result is equivlent to the one tht ou found in prt (). 53. d 54. d d 56. d 3 d d d ( /4 ) + d d / /3 d d 64. d + ( + 3) / () Mke u-substitution (5) to convert the integrnd to rtionl function of u, nd then evlute the integrl. (b) If ou hve CAS, use it to evlute the integrl (no substitution), nd then confirm tht the result is equivlent to tht in prt (). d d sin + cos + sin dθ d cos θ 4 sin 3 cos 69. d sin + tn Use n method to solve for. 7. dt =.5, <<4 t(4 t) 7. t t dt =, > sin sin + tn d Use n method to find the re of the region enclosed b the curves. 73. = 5,=, =, = = 9 4, =, = 75. =,=, =, = = ln, =, = Use n method to find the volume of the solid generted when the region enclosed b the curves is revolved bout the -is. 77. = cos, =, =, = π/ C C 78. = 4, =, = = e,=, =, = 3 8. = ln, =, = Use n method to find the rc length of the curve. 8. =, 8. = 3ln, Use n method to find the re of the surfce generted b revolving the curve bout the -is. 83. = sin, π 84. = /, Informtion is given bout the motion of prticle moving long coordinte line. () Use CAS to find the position function of the prticle for t. (b) Grph the position versus time curve. 85. v(t) = cos 6 t sin 3 t, s() = 86. (t) = e t sin t sin 4t, v() =, s() = FOCUS ON CONCEPTS 87. () Use the substitution u = tn(/) to show tht sec d= ln + tn(/) tn(/) + C nd confirm tht this is consistent with Formul () of Section 7.3. (b) Use the result in prt () to show tht ( π sec d= ln tn 4 + ) + C 88. Use the substitution u = tn(/) to show tht csc d= [ ] cos ln + C + cos nd confirm tht this is consistent with the result in Eercise 65() of Section Find substitution tht cn be used to integrte rtionl functions of sinh nd cosh nd use our substitution to evlute d cosh + sinh without epressing the integrnd in terms of e nd e Some integrls tht cn be evluted b hnd cnnot be evluted b ll computer lgebr sstems. Evlute the integrl b hnd, nd determine if it cn be evluted on our CAS d 8 9. (cos 3 sin 3 cos 3 sin 3 )d 9. 4 d [Hint: ( + ) =?]

121 C d [Hint: Rewrite the denomintor s ( + 9 ).] 94. Let f() = Numericl Integrtion; Simpson s Rule 533 () Use CAS to fctor the denomintor, nd then write down the form of the prtil frction decomposition. You need not find the vlues of the constnts. (b) Check our nswer in prt () b using the CAS to find the prtil frction decomposition of f. (c) Integrte f b hnd, nd then check our nswer b integrting with the CAS. QUICK CHECK ANSWERS 7.6. () Formul (6) (b) Formul (8) (c) Formul () (d) Formul (5). () Formul (5) (b) Formul (5) (c) Formul (8) (d) Formul (97) 3. () 4 ln + + C (b) 6 sin 3 + e sin + C (c) e + sin e + C (d) ln ( ) + tn + C 7.7 NUMERICAL INTEGRATION; SIMPSON S RULE If it is necessr to evlute definite integrl of function for which n ntiderivtive cnnot be found, then one must settle for some kind of numericl pproimtion of the integrl. In Section 5.4 we considered three such pproimtions in the contet of res left endpoint pproimtion, right endpoint pproimtion, nd midpoint pproimtion. In this section we will etend those methods to generl definite integrls, nd we will develop some new methods tht often provide more ccurc with less computtion. We will lso discuss the errors tht rise in integrl pproimtions. A REVIEW OF RIEMANN SUM APPROXIMATIONS Recll from Section 5.5 tht the definite integrl of continuous function f over n intervl [,b] m be computed s b f()d = lim m k n f(k ) k where the sum tht ppers on the right side is clled Riemnn sum. In this formul, k is the width of the kth subintervl of prtition = < < < < n = b of [,b] into n subintervls, nd k denotes n rbitrr point in the kth subintervl. If we tke ll subintervls of the sme width, so tht k = (b )/n, then s n increses the Riemnn sum will eventull be good pproimtion to the definite integrl. We denote this b writing b ( ) b f()d [f( n ) + f( ) + +f( n )] () If we denote the vlues of f t the endpoints of the subintervls b = f(), = f( ), = f( ),..., n = f( n ), n = f( n ) k= nd the vlues of f t the midpoints of the subintervls b m, m,..., mn then it follows from () tht the left endpoint, right endpoint, nd midpoint pproimtions discussed in Section 5.4 cn be epressed s shown in Tble Although we originll

122 534 Chpter 7 / Principles of Integrl Evlution Tble 7.7. left endpoint pproimtion right endpoint pproimtion midpoint pproimtion b f() d b [ n n ] b f() d b [ n n ] b f() d b [ m + m n mn ] n b n n b m m mn m m... m n obtined these results for nonnegtive functions in the contet of pproimting res, the re pplicble to n function tht is continuous on [,b]. n n b Trpezoidl pproimtion Figure 7.7. TRAPEZOIDAL APPROXIMATION It will be convenient in this section to denote the left endpoint, right endpoint, nd midpoint pproimtions with n subintervls b L n, R n, nd M n, respectivel. Of the three pproimtions, the midpoint pproimtion is most widel used in pplictions. If we tke the verge of L n nd R n, then we obtin nother importnt pproimtion denoted b clled the trpezoidl pproimtion: b T n = (L n + R n ) Trpezoidl Approimtion ( ) b f()d T n = [ n + n ] () n The nme trpezoidl pproimtion results from the fct tht in the cse where f is nonnegtive on the intervl of integrtion, the pproimtion T n is the sum of the trpezoidl res shown in Figure 7.7. (see Eercise 5). Emple In Tble 7.7. we hve pproimted ln = d using the midpoint pproimtion nd the trpezoidl pproimtion. used n = subdivisions of the intervl [, ], so tht b = b =. nd } n {{ } n = =.5 }{{ } Midpoint Trpezoidl In ech cse we Throughout this section we will show numericl vlues to nine plces to the right of the deciml point. If our clculting utilit does not show this mn plces, then ou will need to mke the pproprite djustments. Wht is importnt here is tht ou understnd the principles being discussed.

123 7.7 Numericl Integrtion; Simpson s Rule 535 midpoint pproimtion Tble 7.7. midpoint nd trpezoidl pproimtions for d trpezoidl pproimtion i midpoint m i mi = f (m i ) = /m i i endpoint i i = f ( i ) = / i multiplier w i w i i d (.)( ) d (.5)( ) B rewriting (3) nd (4) in the form b f()d = pproimtion + error we see tht positive vlues of E M nd E T correspond to underestimtes nd negtive vlues to overestimtes. COMPARISON OF THE MIDPOINT AND TRAPEZOIDAL APPROXIMATIONS We define the errors in the midpoint nd trpezoidl pproimtions to be E M = b f()d M n nd E T = b f()d T n (3 4) respectivel, nd we define E M nd E T to be the bsolute errors in these pproimtions. The bsolute errors re nonnegtive nd do not distinguish between underestimtes nd overestimtes. Emple The vlue of ln to nine deciml plces is ln = d (5) so we see from Tbles 7.7. nd tht the bsolute errors in pproimting ln b M nd T re E M = ln M.38 E T = ln T.64 Thus, the midpoint pproimtion is more ccurte thn the trpezoidl pproimtion in this cse. Tble ln (nine deciml plces) pproimtion M T error E M = ln M.38 E T = ln T.64

124 536 Chpter 7 / Principles of Integrl Evlution Figure 7.7. m k The shded tringles hve equl res. It is not ccidentl in Emple tht the midpoint pproimtion of ln ws more ccurte thn the trpezoidl pproimtion. To see wh this is so, we first need to look t the midpoint pproimtion from nother point of view. To simplif our eplntion, we will ssume tht f is nonnegtive on [,b], though the conclusions we rech will be true without this ssumption. If f is differentible function, then the midpoint pproimtion is sometimes clled the tngent line pproimtion becuse for ech subintervl of [,b] the re of the rectngle used in the midpoint pproimtion is equl to the re of the trpezoid whose upper boundr is the tngent line to = f() t the midpoint of the subintervl (Figure 7.7.). The equlit of these res follows from the fct tht the shded res in the figure re congruent. We will now show how this point of view bout midpoint pproimtions cn be used to estblish useful criteri for determining which of M n or T n produces the better pproimtion of given integrl. In Figure we hve isolted subintervl of [,b] on which the grph of function f is concve down, nd we hve shded the res tht represent the errors in the midpoint nd trpezoidl pproimtions over the subintervl. In Figure 7.7.3b we show succession of four illustrtions which mke it evident tht the error from the midpoint pproimtion is less thn tht from the trpezoidl pproimtion. If the grph of f were concve up, nlogous figures would led to the sme conclusion. (This rgument, due to Frnk Buck, ppered in The College Mthemtics Journl, Vol. 6, No., 985.) Midpoint error Trpezoidl error Justif the conclusions in ech step of Figure 7.7.3b. m k () Figure Blue re < Blue re = Blue re < Yellow re (b) Figure lso suggests tht on subintervl where the grph is concve down, the midpoint pproimtion is lrger thn the vlue of the integrl nd the trpezoidl pproimtion is smller. On n intervl where the grph is concve up it is the other w round. In summr, we hve the following result, which we stte without forml proof: 7.7. theorem Let f be continuous on [,b], nd let E M nd E T be the bsolute errors tht result from the midpoint nd trpezoidl pproimtions of b f()d using n subintervls. () If the grph of f is either concve up or concve down on (, b), then E M < E T, tht is, the bsolute error from the midpoint pproimtion is less thn tht from the trpezoidl pproimtion. (b) (c) If the grph of f is concve down on (, b), then T n < b f()d < M n If the grph of f is concve up on (, b), then M n < b f()d < T n

125 7.7 Numericl Integrtion; Simpson s Rule 537 WARNING Do not conclude tht the midpoint pproimtion is lws better thn the trpezoidl pproimtion; the trpezoidl pproimtion m be better if the function chnges concvit on the intervl of integrtion. Emple 3 Since the grph of f() = / is continuous on the intervl [, ] nd concve up on the intervl (, ), it follows from prt () of Theorem 7.7. tht M n will lws provide better pproimtion thn T n for d = ln Moreover, if follows from prt (c) of Theorem 7.7. tht M n < ln <T n for ever positive integer n. Note tht this is consistent with our computtions in Emple. Emple 4 The midpoint nd trpezoidl pproimtions cn be used to pproimte sin b using the integrl sin = cos d Since f() = cos is continuous on [, ] nd concve down on (, ), it follows from prts () nd (b) of Theorem 7.7. tht the bsolute error in M n will be less thn tht in T n, nd tht T n < sin <M n for ever positive integer n. This is consistent with the results in Tble for n = 5 (intermedite computtions re omitted). sin (nine deciml plces) Tble pproimtion M T error E M = sin M E T = sin T Emple 5 Tble shows pproimtions for sin 3 = 3 cos dusing the midpoint nd trpezoidl pproimtions with n = subdivisions of the intervl [, 3]. Note tht E M < E T nd T < sin 3 <M, lthough these results re not gurnteed b Theorem 7.7. since f() = cos chnges concvit on the intervl [, 3]. Tble sin 3 (nine deciml plces) pproimtion M.4656 T.467 error E M = sin 3 M.5359 E T = sin 3 T.5999 SIMPSON S RULE When the left nd right endpoint pproimtions re verged to produce the trpezoidl pproimtion, better pproimtion often results. We now see how weighted verge of the midpoint nd trpezoidl pproimtions cn ield n even better pproimtion. The numericl evidence in Tbles 7.7.3, 7.7.4, nd revels tht E T E M,so tht E M + E T in these instnces. This suggests tht 3 b f()d = b f()d+ b f()d = (M k + E M ) + (T k + E T ) = (M k + T k ) + (E M + E T ) M k + T k

126 538 Chpter 7 / Principles of Integrl Evlution WARNING Note tht in (7) the subscript n in S n is lws even since it is twice the vlue of the subscripts for the corresponding midpoint nd trpezoidl pproimtions. For emple, nd S = 3 (M 5 + T 5 ) S = 3 (M + T ) This gives b f()d 3 (M k + T k ) (6) The midpoint pproimtion M k in (6) requires the evlution of f t k points in the intervl [,b], nd the trpezoidl pproimtion T k in (6) requires the evlution of f t k + points in [,b]. Thus, 3 (M k + T k ) uses k + vlues of f, tken t equll spced points in the intervl [,b]. These points re obtined b prtitioning [,b] into k equl subintervls indicted b the left endpoints, right endpoints, nd midpoints used in T k nd M k, respectivel. Setting n = k, we use S n to denote the weighted verge of M k nd T k in (6). Tht is, S n = S k = 3 (M k + T k ) or S n = 3 (M n/ + T n/ ) (7) Tble displs the pproimtions S n corresponding to the dt in Tbles to Tble function vlue (nine deciml plces) pproimtion error ln sin sin 3.48 (/) d S = (M + T ) cos d S = (M 5 + T 5 ) cos d S = (M + T ) Using the midpoint pproimtion formul in Tble 7.7. nd Formul () for the trpezoidl pproimtion, we cn derive similr formul for S n. We strt b prtitioning the intervl [,b] into n even number of equl subintervls. If n is the number of subintervls, then ech subintervl hs length (b )/n. Lbel the endpoints of these subintervls successivel b =,,,..., n = b. Then,, 4,..., n define prtition of [,b] into n/ equl intervls, ech of length (b )/n, nd the midpoints of these subintervls re, 3, 5,..., n, respectivel, s illustrted in Figure Using i = f( i ),we hve ( ) (b ) M n/ = [ n ] n ( ) b = [ n ] n Noting tht (b )/[(n/)] =(b )/n, we cn epress T n/ s ( ) b T n/ = [ n + n ] n Thus, S n = 3 (M n/ + T n/ ) cn be epressed s S n = ( ) b [ n + 4 n + n ] (8) 3 n The pproimtion b f()d S n (9) with S n s given in (8) is known s Simpson s rule. We denote the error in this pproimtion b b E S = f()d S n () As before, the bsolute error in the pproimtion (9) is given b E S.

127 7.7 Numericl Integrtion; Simpson s Rule 539 n n = b = n Figure midpoint midpoint midpoint midpoint... n n n n Emple 6 In Tble we hve used Simpson s rule with n = subintervls to obtin the pproimtion For this pproimtion, ln = 3 d S = ( ) b = ( ) n 3 = 3 i endpoint i Tble n pproimtion to ln using simpson's rule i = f( i ) = / i multiplier w i w i i d 3 ( ) Thoms Simpson (7 76) English mthemticin. Simpson ws the son of wever. He ws trined to follow in his fther s footsteps nd hd little forml eduction in his erl life. His interest in science nd mthemtics ws roused in 74, when he witnessed n eclipse of the Sun nd received two books from peddler, one on strolog nd the other on rithmetic. Simpson quickl bsorbed their contents nd soon becme successful locl fortune teller. His improved finncil sitution enbled him to give up weving nd mrr his lndld. Then in 733 some msterious unfortunte incident forced him to move. He settled in Derb, where he tught in n evening school nd worked t weving during the d. In 736 he moved to London nd published his first mthemticl work in periodicl clled the Ldies Dir (of which he lter becme the editor). In 737 he published successful clculus tetbook tht enbled him to give up weving completel nd concentrte on tetbook writing nd teching. His fortunes improved further in 74 when one Robert Heth ccused him of plgirism. The publicit ws mrvelous, nd Simpson proceeded to dsh off succession of best-selling tetbooks: Algebr (ten editions plus trnsltions), Geometr (twelve editions plus trnsltions), Trigonometr (five editions plus trnsltions), nd numerous others. It is interesting to note tht Simpson did not discover the rule tht bers his nme it ws well-known result b Simpson s time. [Imge:

128 54 Chpter 7 / Principles of Integrl Evlution Although S is weighted verge of M 5 nd T 5, it mkes sense to compre S to M nd T, since the sums for these three pproimtions involve the sme number of terms. Using the vlues for M nd T from Emple nd the vlue for S in Tble 7.7.7, we hve E M = ln M =.38 E T = ln T =.64 E S = ln S =.35 Compring these bsolute errors, it is cler tht S is much more ccurte pproimtion of ln thn either M or T. = f() = A + B + C Y Y Y m Δ Δ Figure GEOMETRIC INTERPRETATION OF SIMPSON S RULE The midpoint (or tngent line) pproimtion nd the trpezoidl pproimtion for definite integrl re bsed on pproimting segment of the curve = f()b line segments. Intuition suggests tht we might improve on these pproimtions using prbolic rcs rther thn line segments, thereb ccounting for concvit of the curve = f() more closel. At the hert of this ide is formul, sometimes clled the one-third rule. The one-third rule epresses definite integrl of qudrtic function g() = A + B + C in terms of the vlues Y, Y, nd Y of g t the left endpoint, midpoint, nd right endpoint, respectivel, of the intervl of integrtion [m, m + ] (see Figure 7.7.5): m+ m (A + B + C)d = 3 [Y + 4Y + Y ] () Verifiction of the one-third rule is left for the reder (Eercise 53). B ppling the one-third rule to subintervls [ k, k ],k =,...,n/, one rrives t Formul (8) for Simpson s rule (Eercise 54). Thus, Simpson s rule corresponds to the integrl of piecewise-qudrtic pproimtion to f(). ERROR BOUNDS With ll the methods studied in this section, there re two sources of error: the intrinsic or trunction error due to the pproimtion formul, nd the roundoff error introduced in the clcultions. In generl, incresing n reduces the trunction error but increses the roundoff error, since more computtions re required for lrger n. In prcticl pplictions, it is importnt to know how lrge n must be tken to ensure tht specified degree of ccurc is obtined. The nlsis of roundoff error is complicted nd will not be considered here. However, the following theorems, which re proved in books on numericl nlsis, provide upper bounds on the trunction errors in the midpoint, trpezoidl, nd Simpson s rule pproimtions theorem (Midpoint nd Trpezoidl Error Bounds) If f is continuous on [,b] nd if K is the mimum vlue of f () on [,b], then b () E M = f()d M n (b )3 K 4n () b (b) E T = f()d T n (b )3 K n (3)

129 7.7 Numericl Integrtion; Simpson s Rule theorem (Simpson Error Bound) If f (4) is continuous on [,b] nd if K 4 is the mimum vlue of f (4) () on [,b], then b E S = f()d S n (b )5 K 4 (4) 8n 4 Emple 7 Find n upper bound on the bsolute error tht results from pproimting ln = d using () the midpoint pproimtion M, (b) the trpezoidl pproimtion T, nd (c) Simpson s rule S, ech with n = subintervls. Solution. We will ppl Formuls (), (3), nd (4) with f() =, =, b =, nd n = Note tht the upper bounds clculted in Emple 7 re consistent with the vlues E M, E T, nd E S clculted in Emple 6 but re considerbl greter thn those vlues. It is quite common tht the upper bounds on the bsolute errors given in Theorems 7.7. nd substntill eceed the ctul bsolute errors. However, tht does not diminish the utilit of these bounds. We hve Thus, f () =, f () = 3, f () = 6 4, f(4) () = 4 f () = =, f (4) () = 4 3 = 4 3 where we hve dropped the bsolute vlues becuse f () nd f (4) () hve positive vlues for. Since f () nd f 4 () re continuous nd decresing on [, ], both functions hve their mimum vlues t = ; for f () this mimum vlue is nd for f 4 () the mimum vlue is 4. Thus we cn tke K = in () nd (3), nd K 4 = 4 in (4). This ields E M (b )3 K 4n = E T (b )3 K n = E S (b )5 K 4 8n 4 = Emple 8 How mn subintervls should be used in pproimting ln = d b Simpson s rule for five deciml-plce ccurc? Solution. To obtin five deciml-plce ccurc, we must choose the number of subintervls so tht E S.5 = 5 6 From (4), this cn be chieved b tking n in Simpson s rule to stisf (b ) 5 K 4 8n 4 5 6

130 54 Chpter 7 / Principles of Integrl Evlution Tking =,b =, nd K 4 = 4 (found in Emple 7) in this inequlit ields n which, on tking reciprocls, cn be rewritten s Thus, n = n Since n must be n even integer, the smllest vlue of n tht stisfies this requirement is n = 4. Thus, the pproimtion S 4 using 4 subintervls will produce five deciml-plce ccurc. REMARK In cses where it is difficult to find the vlues of K nd K 4 in Formuls (), (3), nd (4), these constnts m be replced b n lrger constnts. For emple, suppose tht constnt K cn be esil found with the certint tht f () <Kon the intervl. Then K K nd E T (b )3 K n (b )3 K n (5) so the right side of (5) is lso n upper bound on the vlue of E T. Using K, however, will likel increse the computed vlue of n needed for given error tolernce. Mn pplictions involve the resolution of competing prcticl issues, illustrted here through the trde-off between the convenience of finding crude bound for f () versus the efficienc of using the smllest possible n for desired ccurc. Emple 9 How mn subintervls should be used in pproimting cos( )d b the midpoint pproimtion for three deciml-plce ccurc? Solution. To obtin three deciml-plce ccurc, we must choose n so tht E M.5 = 5 4 (6) From () with f() = cos( ), =, nd b =, n upper bound on E M is given b 4 3 = f () = 4 cos( ) + sin( ) Figure E M K (7) 4n where K is the mimum vlue of f () on the intervl [, ]. However, so tht f () = sin( ) f () = 4 cos( ) sin( ) = [4 cos( ) + sin( )] f () = 4 cos( ) + sin( ) (8) It would be tedious to look for the mimum vlue of this function on the intervl [, ]. For in [, ], it is es to see tht ech of the epressions, cos( ), nd sin( ) is bounded in bsolute vlue b, so 4 cos( ) + sin( ) 4 + = 6on[, ]. We cn improve on this b using grphing utilit to sketch f (), s shown in Figure It is evident from the grph tht f () < 4 for

131 Thus, it follows from (7) tht E M K 4n < 4 4n = 6n nd hence we cn stisf (6) b choosing n so tht 6n < 5 4 which, on tking reciprocls, cn be written s n > Numericl Integrtion; Simpson s Rule 543 or n> The smllest integer vlue of n stisfing this inequlit is n = 9. Thus, the midpoint pproimtion M 9 using 9 subintervls will produce three deciml-plce ccurc. A COMPARISON OF THE THREE METHODS Of the three methods studied in this section, Simpson s rule generll produces more ccurte results thn the midpoint or trpezoidl pproimtions for the sme mount of work. To mke this plusible, let us epress (), (3), nd (4) in terms of the subintervl width = b n We obtin E M 4 K (b )( ) (9) E T K (b )( ) () E S 8 K 4(b )( ) 4 () (verif). For Simpson s rule, the upper bound on the bsolute error is proportionl to ( ) 4, wheres the upper bound on the bsolute error for the midpoint nd trpezoidl pproimtions is proportionl to ( ). Thus, reducing the intervl width b fctor of, for emple, reduces the error bound b fctor of for the midpoint nd trpezoidl pproimtions but reduces the error bound b fctor of, for Simpson s rule. This suggests tht, s n increses, the ccurc of Simpson s rule improves much more rpidl thn tht of the other pproimtions. As finl note, observe tht if f() is polnomil of degree 3 or less, then we hve f (4) () = for ll, sok 4 = in (4) nd consequentl E S =. Thus, Simpson s rule gives ect results for polnomils of degree 3 or less. Similrl, the midpoint nd trpezoidl pproimtions give ect results for polnomils of degree or less. (You should lso be ble to see tht this is so geometricll.) QUICK CHECK EXERCISES 7.7 (See pge 547 for nswers.). Let T n be the trpezoidl pproimtion for the definite integrl of f() over n intervl [,b] using n subintervls. () Epressed in terms of L n nd R n (the left nd right endpoint pproimtions), T n =. (b) Epressed in terms of the function vlues,,..., n t the endpoints of the subintervls, T n =.. Let I denote the definite integrl of f over n intervl [,b] with T n nd M n the respective trpezoidl nd midpoint pproimtions of I for given n. Assume tht the grph of f is concve up on the intervl [,b] nd order the quntities T n,m n, nd I from smllest to lrgest: < <.

132 544 Chpter 7 / Principles of Integrl Evlution 3. Let S 6 be the Simpson s rule pproimtion for b f()d using n = 6 subintervls. () Epressed in terms of M 3 nd T 3 (the midpoint nd trpezoidl pproimtions), S 6 =. (b) Using the function vlues,,,..., 6 t the endpoints of the subintervls, S 6 =. 4. Assume tht f (4) is continuous on [, ] nd tht f (k) () stisfies f (k) () on[, ],k =,, 3, 4. Find n upper bound on the bsolute error tht results from pproimting the integrl of f over [, ] using () the midpoint pproimtion M ; (b) the trpezoidl pproimtion T ; nd (c) Simpson s rule S Approimte d using the indicted method. () M = (b) T = (c) S = EXERCISE SET 7.7 C CAS 6 Approimte the integrl using () the midpoint pproimtion M, (b) the trpezoidl pproimtion T, nd (c) Simpson s rule pproimtion S using Formul (7). In ech cse, find the ect vlue of the integrl nd pproimte the bsolute error. Epress our nswers to t lest four deciml plces. 3 9 π/. + d. d 3. cos d 4. sin d e d d 7 Use inequlities (), (3), nd (4) to find upper bounds on the errors in prts (), (b), nd (c) of the indicted eercise. 7. Eercise 8. Eercise 9. Eercise 3. Eercise 4. Eercise 5. Eercise Use inequlities (), (3), nd (4) to find number n of subintervls for () the midpoint pproimtion M n, (b) the trpezoidl pproimtion T n, nd (c) Simpson s rule pproimtion S n to ensure tht the bsolute error will be less thn the given vlue. 3. Eercise ; Eercise ; Eercise 3; 3 6. Eercise 4; 3 7. Eercise 5; 4 8. Eercise 6; 4 9 True Flse Determine whether the sttement is true or flse. Eplin our nswer. 9. The midpoint pproimtion, M n, is the verge of the left nd right endpoint pproimtions, L n nd R n, respectivel.. If f() is concve down on the intervl (, b), then the trpezoidl pproimtion T n underestimtes b f()d.. The Simpson s rule pproimtion S 5 for b f()d is weighted verge of the pproimtions M 5 nd T 5, where M 5 is given twice the weight of T 5 in the verge.. Simpson s rule pproimtion S 5 for b f()d corresponds to b q()d, where the grph of q is composed of 5 prbolic segments joined t points on the grph of f. 3 4 Find function g() of the form g() = A + B + C whose grph contins the points (m, f(m )), (m, f(m)), nd (m +, f(m + )), for the given function f() nd the given vlues of m nd. Then verif Formul (): m+ g() d = m 3 [Y + 4Y + Y ] where Y = f(m ), Y = f(m), nd Y = f(m + ). 3. f() = ; m = 3, = 4. f() = sin (π); m = 6, = Approimte the integrl using Simpson s rule S nd compre our nswer to tht produced b clculting utilit with numericl integrtion cpbilit. Epress our nswers to t lest four deciml plces e d d d 8. cos( )d 3. π + sin d (ln ) 3/ d 3 3 The ect vlue of the given integrl is π (verif). Approimte the integrl using () the midpoint pproimtion M, (b) the trpezoidl pproimtion T, nd (c) Simpson s rule pproimtion S using Formul (7). Approimte the bsolute error nd epress our nswers to t lest four deciml plces d 3. 9 d In Emple 8 we showed tht tking n = 4 subdivisions ensures tht the pproimtion of ln = d b Simpson s rule is ccurte to five deciml plces. Confirm this b compring the pproimtion of ln produced b Simpson s rule with n = 4 to the vlue produced directl b our clculting utilit.

133 7.7 Numericl Integrtion; Simpson s Rule In ech prt, determine whether trpezoidl pproimtion would be n underestimte or n overestimte for the definite integrl. () cos( )d (b) 3/ cos( )d Find vlue of n to ensure tht the bsolute error in pproimting the integrl b the midpoint pproimtion will be less thn 4. Estimte the bsolute error, nd epress our nswers to t lest four deciml plces. 35. sin d 36. e cos d Show tht the inequlities () nd (3) re of no vlue in finding n upper bound on the bsolute error tht results from pproimting the integrl using either the midpoint pproimtion or the trpezoidl pproimtion. 37. d 38. sin d 39 4 Use Simpson s rule pproimtion S to pproimte the length of the curve over the stted intervl. Epress our nswers to t lest four deciml plces. 39. = sin from = to = π 4. = from = to = FOCUS ON CONCEPTS 4. A grph of the speed v versus time t curve for test run of BMW 335i is shown in the ccompning figure. Estimte the speeds t times t =, 5,, 5,, 5, 3 s from the grph, convert to ft/s using mi/h = 5 ft/s, nd use these speeds nd Simpson s rule to pproimte the number of feet trveled during the first 3 s. Round our nswer to the nerest foot. [Hint: Distnce trveled = 3 v(t) dt.] Source: Dt from Cr nd Driver Mgzine, September. Speed v (mi/h) Time t (s) Figure E-4 4. A grph of the ccelertion versus time t for n object moving on stright line is shown in the ccompning figure. Estimte the ccelertions t t =,,,...,8 seconds (s) from the grph nd use Simpson s rule to pproimte the chnge in velocit from t = tot = 8s. Round our nswer to the nerest tenth cm/s. [Hint: Chnge in velocit = 8 (t)dt.] Accelertion (cm/s ) Time t (s) Figure E Numericl integrtion methods cn be used in problems where onl mesured or eperimentll determined vlues of the integrnd re vilble. Use Simpson s rule to estimte the vlue of the relevnt integrl in these eercises. 43. The ccompning tble gives the speeds, in miles per second, t vrious times for test rocket tht ws fired upwrd from the surfce of the Erth. Use these vlues to pproimte the number of miles trveled during the first 8 s. Round our nswer to the nerest tenth of mile. [Hint: Distnce trveled = 8 v(t) dt.] time t (s) speed v (mi/s) Tble E The ccompning tble gives the speeds of bullet t vrious distnces from the muzzle of rifle. Use these vlues to pproimte the number of seconds for the bullet to trvel 8 ft. Epress our nswer to the nerest hundredth of second. [Hint: If v is the speed of the bullet nd is the distnce trveled, then v = d/dt so tht dt/d = /v nd t = 8 (/v) d.] distnce (ft) speed v (ft/s) Tble E Mesurements of potter shrd recovered from n rcheologicl dig revel tht the shrd cme from pot with flt bottom nd circulr cross sections (see the ccompning

134 546 Chpter 7 / Principles of Integrl Evlution figure below). The figure shows interior rdius mesurements of the shrd mde ever 4 cm from the bottom of the pot to the top. Use those vlues to pproimte the interior volume of the pot to the nerest tenth of liter ( L = cm 3 ). [Hint: Use 6..3 (volume b cross sections) to set up n pproprite integrl for the volume.] (cm) 6.8 cm 5.4 cm 3.8 cm.5 cm 8.5 cm Figure E Engineers wnt to construct stright nd level rod 6 ft long nd 75 ft wide b mking verticl cut through n intervening hill (see the ccompning figure). Heights of the hill bove the centerline of the proposed rod, s obtined t vrious points from contour mp of the region, re shown in the ccompning figure. To estimte the construction costs, the engineers need to know the volume of erth tht must be removed. Approimte this volume, rounded to the nerest cubic foot. [Hint: First set up n integrl for the cross-sectionl re of the cut long the centerline of the rod, then ssume tht the height of the hill does not vr between the centerline nd edges of the rod.] horizontl distnce (ft) Figure E-46 height h (ft) Centerline 6 ft 75 ft C 47. Let f() = cos( ). () Use CAS to pproimte the mimum vlue of f () on the intervl [, ]. (b) How lrge must n be in the midpoint pproimtion of f()d to ensure tht the bsolute error is less thn 5 4? Compre our result with tht obtined in Emple 9. (c) Estimte the integrl using the midpoint pproimtion with the vlue of n obtined in prt (b). C 48. Let f() = + 3. () Use CAS to pproimte the mimum vlue of f () on the intervl [, ]. (b) How lrge must n be in the trpezoidl pproimtion of f()d to ensure tht the bsolute error is less thn 3? (c) Estimte the integrl using the trpezoidl pproimtion with the vlue of n obtined in prt (b). C 49. Let f() = cos( ). () Use CAS to pproimte the mimum vlue of f (4) () on the intervl [, ]. (b) How lrge must the vlue of n be in the pproimtion S n of f()d b Simpson s rule to ensure tht the bsolute error is less thn 4? (c) Estimte the integrl using Simpson s rule pproimtion S n with the vlue of n obtined in prt (b). C 5. Let f() = + 3. () Use CAS to pproimte the mimum vlue of f (4) () on the intervl [, ]. (b) How lrge must the vlue of n be in the pproimtion S n of f()d b Simpson s rule to ensure tht the bsolute error is less thn 6? (c) Estimte the integrl using Simpson s rule pproimtion S n with the vlue of n obtined in prt (b). FOCUS ON CONCEPTS 5. () Verif tht the verge of the left nd right endpoint pproimtions s given in Tble 7.7. gives Formul () for the trpezoidl pproimtion. (b) Suppose tht f is continuous nonnegtive function on the intervl [,b] nd prtition [,b] with equll spced points, = < < < n = b. Find the re of the trpezoid under the line segment joining points ( k,f( k )) nd ( k+,f( k+ )) nd bove the intervl [ k, k+ ]. Show tht the right side of Formul () is the sum of these trpezoidl res (Figure 7.7.). 5. Let f be function tht is positive, continuous, decresing, nd concve down on the intervl [,b]. Assuming tht [,b] is subdivided into n equl subintervls, rrnge the following pproimtions of b f()d in order of incresing vlue: left endpoint, right endpoint, midpoint, nd trpezoidl. 53. Suppose tht > nd g() = A + B + C. Let m be number nd set Y = g(m ), Y = g(m), nd Y = g(m + ). Verif Formul (): m+ g() d = 3 [Y + 4Y + Y ] m 54. Suppose tht f is continuous nonnegtive function on the intervl [,b], n is even, nd [,b] is prtitioned using n + equll spced points, = < < < n = b. Set = f( ), = f( ),..., n = f( n ). Let g,g,...,g n/ be the qudrtic functions of the form g i () = A + B + C so tht (cont.)

135 7.8 Improper Integrls 547 the grph of g psses through the points (, ), (, ), nd (, ); the grph of g psses through the points (, ), ( 3, 3 ), nd ( 4, 4 );... the grph of g n/ psses through the points ( n, n ), ( n, n ), nd ( n, n ). Verif tht Formul (8) computes the re under piecewise qudrtic function b showing tht ( n/ ) j g j () d j= j = ( ) b [ n + n + 4 n + n ] 55. Writing Discuss two different circumstnces under which numericl integrtion is necessr. 56. Writing For the numericl integrtion methods of this section, better ccurc of n pproimtion ws obtined b incresing the number of subdivisions of the intervl. Another strteg is to use the sme number of subintervls, but to select subintervls of differing lengths. Discuss scheme for doing this to pproimte 4 dusing trpezoidl pproimtion with 4 subintervls. Comment on the dvntges nd disdvntges of our scheme. QUICK CHECK ANSWERS 7.7 ( b. () (L n + R n ) (b) ( n b (b) 8 ) [ n + n ]. M n <I<T n 3. () 3 M T 3 ) ( ) 4. () 5. () M = (b) T = 9 (c) S = (b) (c),8, 7.8 IMPROPER INTEGRALS Up to now we hve focused on definite integrls with continuous integrnds nd finite intervls of integrtion. In this section we will etend the concept of definite integrl to include infinite intervls of integrtion nd integrnds tht become infinite within the intervl of integrtion. IMPROPER INTEGRALS It is ssumed in the definition of the definite integrl b f()d tht [,b] is finite intervl nd tht the limit tht defines the integrl eists; tht is, the function f is integrble. We observed in Theorems 5.5. nd tht continuous functions re integrble, s re bounded functions with finitel mn points of discontinuit. We lso observed in Theorem tht functions tht re not bounded on the intervl of integrtion re not integrble. Thus, for emple, function with verticl smptote within the intervl of integrtion would not be integrble. Our min objective in this section is to etend the concept of definite integrl to llow for infinite intervls of integrtion nd integrnds with verticl smptotes within the intervl of integrtion. We will cll the verticl smptotes infinite discontinuities, nd we will cll

136 548 Chpter 7 / Principles of Integrl Evlution integrls with infinite intervls of integrtion or infinite discontinuities within the intervl of integrtion improper integrls. Here re some emples: Improper integrls with infinite intervls of integrtion: + d +, e d d, + Improper integrls with infinite discontinuities in the intervl of integrtion: 3 d 3, d π, tn d Improper integrls with infinite discontinuities nd infinite intervls of integrtion: + d, + d 9, + sec d Figure 7.8. = Figure 7.8. Are = + f () d b b d = b INTEGRALS OVER INFINITE INTERVALS To motivte resonble definition for improper integrls of the form + f()d let us begin with the cse where f is continuous nd nonnegtive on [,+ ), so we cn think of the integrl s the re under the curve = f()over the intervl [,+ ) (Figure 7.8.). At first, ou might be inclined to rgue tht this re is infinite becuse the region hs infinite etent. However, such n rgument would be bsed on vgue intuition rther thn precise mthemticl logic, since the concept of re hs onl been defined over intervls of finite etent. Thus, before we cn mke n resonble sttements bout the re of the region in Figure 7.8., we need to begin b defining wht we men b the re of this region. For tht purpose, it will help to focus on specific emple. Suppose we re interested in the re A of the region tht lies below the curve = / nd bove the intervl [, + ) on the -is. Insted of tring to find the entire re t once, let us begin b clculting the portion of the re tht lies bove finite intervl [,b], where b>is rbitrr. Tht re is b d = = b (Figure 7.8.). If we now llow b to increse so tht b +, then the portion of the re over the intervl [,b] will begin to fill out the re over the entire intervl [, + ) (Figure 7.8.3), nd hence we cn resonbl define the re A under = / over the intervl [, + ) to be + d b A = = lim d b + Thus, the re hs finite vlue of nd is not infinite s we first conjectured. ] b ( = lim ) = () b + b = = = = Figure Are = 3 Are = 3 Are = Are = With the preceding discussion s our guide, we mke the following definition (which is pplicble to functions with both positive nd negtive vlues).

137 7.8 Improper Integrls 549 If f is nonnegtive over the intervl [,+ ), then the improper integrl in Definition 7.8. cn be interpreted to be the re under the grph of f over the intervl [,+ ). If the integrl converges, then the re is finite nd equl to the vlue of the integrl, nd if the integrl diverges, then the re is regrded to be infinite definition The improper integrl of f over the intervl [, + ) is defined to be + f()d = lim b + b f()d In the cse where the limit eists, the improper integrl is sid to converge, nd the limit is defined to be the vlue of the integrl. In the cse where the limit does not eist, the improper integrl is sid to diverge, nd it is not ssigned vlue. Emple Evlute () + d 3 (b) + d Solution (). Following the definition, we replce the infinite upper limit b finite upper limit b, nd then tke the limit of the resulting integrl. This ields + d b [ = lim d 3 b + = lim ] b ( = lim 3 b + b + ) = b Since the limit is finite, the integrl converges nd its vlue is /. = 3 = = Figure Solution (b). + d b = lim d b + = lim [ ] b ln b + = lim ln b =+ b + In this cse the integrl diverges nd hence hs no vlue. Becuse the functions / 3, /, nd / re nonnegtive over the intervl [, + ), it follows from () nd the lst emple tht over this intervl the re under = / 3 is, the re under = / is, nd the re under = / is infinite. However, on the surfce the grphs of the three functions seem ver much like (Figure 7.8.4), nd there is nothing to suggest wh one of the res should be infinite nd the other two finite. One eplntion is tht / 3 nd / pproch zero more rpidl thn / s +, so tht the re over the intervl [,b] ccumultes less rpidl under the curves = / 3 nd = / thn under = / s b +, nd the difference is just enough tht the first two res re finite nd the third is infinite. + d Emple For wht vlues of p does the integrl converge? p Solution. We know from the preceding emple tht the integrl diverges if p =, so let us ssume tht p =. In this cse we hve + d b = lim p p ] b [ b p d = lim = lim p b + b + p b + p ] p If p>, then the eponent p is negtive nd b p sb + ; nd if p<, then the eponent p is positive nd b p + s b +. Thus, the integrl converges if p>nd diverges otherwise. In the convergent cse the vlue of the integrl is + d p = [ p ] = p (p > )

138 55 Chpter 7 / Principles of Integrl Evlution The following theorem summrizes this result theorem + d p = if p> p diverges if p Emple 3 Evlute + ( )e d. Solution. We begin b evluting the indefinite integrl using integrtion b prts. Setting u = nd dv = e d ields ( )e d = e ( ) e d = e + e + e + C = e + C = ( )e 3 The net signed re between the grph nd the intervl [, + ) is zero. Figure If f is nonnegtive over the intervl (, + ), then the improper integrl + f()d cn be interpreted to be the re under the grph of f over the intervl (, + ). The re is finite nd equl to the vlue of the integrl if the integrl converges nd is infinite if it diverges. Thus, + ( )e d = lim b + b ( )e d = lim [e ] b = lim b b + b + e b The limit is n indeterminte form of tpe /, so we will ppl L Hôpitl s rule b differentiting the numertor nd denomintor with respect to b. This ields + ( )e d = lim b + e = b We cn interpret this to men tht the net signed re between the grph of = ( )e nd the intervl [, + ) is (Figure 7.8.5) definition The improper integrl of f over the intervl (,b] is defined to be b b f()d = f()d () lim The integrl is sid to converge if the limit eists nd diverge if it does not. The improper integrl of f over the intervl (, + ) is defined s + f()d = c f()d + + c f()d (3) where c is n rel number. The improper integrl is sid to converge if both terms converge nd diverge if either term diverges. Emple 4 + d Evlute +. Although we usull choose c = in (3), the choice does not mtter becuse it cn be proved tht neither the convergence nor the vlue of the integrl is ffected b the choice of c. Solution. we obtin + d + = lim b + d + = lim We will evlute the integrl b choosing c = in (3). With this vlue for c b d [ ] b + = lim tn = lim b + b + (tn b) = π d + = lim [ tn ] = lim ( tn ) = π

139 Are = c = + Thus, the integrl converges nd its vlue is + d + = d Improper Integrls 55 d + = π + π = π Since the integrnd is nonnegtive on the intervl (, + ), the integrl represents the Figure re of the region shown in Figure b f() d b INTEGRALS WHOSE INTEGRANDS HAVE INFINITE DISCONTINUITIES Net we will consider improper integrls whose integrnds hve infinite discontinuities. We will strt with the cse where the intervl of integrtion is finite intervl [,b] nd the infinite discontinuit occurs t the right-hnd endpoint. To motivte n pproprite definition for such n integrl let us consider the cse where f is nonnegtive on [,b], so we cn interpret the improper integrl b f()d s the re of the region in Figure The problem of finding the re of this region is complicted b the fct tht it etends indefinitel in the positive -direction. However, insted of tring to find the entire re t once, we cn proceed indirectl b clculting the portion of the re over the intervl [,k], where k<b, nd then letting k pproch b to fill out the re of the entire region (Figure 7.8.7b). Motivted b this ide, we mke the following definition. () definition If f is continuous on the intervl [,b], ecept for n infinite discontinuit t b, then the improper integrl of f over the intervl [, b] is defined s k f() d b f()d = lim k b k f()d (4) k b (b) Figure = Figure In the cse where the indicted limit eists, the improper integrl is sid to converge, nd the limit is defined to be the vlue of the integrl. In the cse where the limit does not eist, the improper integrl is sid to diverge, nd it is not ssigned vlue. Emple 5 d Evlute. Solution. The integrl is improper becuse the integrnd pproches + s pproches the upper limit from the left (Figure 7.8.8). From (4), d k = lim k d = lim [ = lim k + k k [ ] = ] k Improper integrls with n infinite discontinuit t the left-hnd endpoint or inside the intervl of integrtion re defined s follows.

140 55 Chpter 7 / Principles of Integrl Evlution definition If f is continuous on the intervl [,b], ecept for n infinite discontinuit t, then the improper integrl of f over the intervl [, b] is defined s b f()d = lim k + b k f()d (5) The integrl is sid to converge if the indicted limit eists nd diverge if it does not. If f is continuous on the intervl [,b], ecept for n infinite discontinuit t point c in (, b), then the improper integrl of f over the intervl [, b] is defined s c b f() d f() d c c b b f()d = c f()d + b c f()d (6) where the two integrls on the right side re themselves improper. The improper integrl on the left side is sid to converge if both terms on the right side converge nd diverge if either term on the right side diverges (Figure 7.8.9). b f() d is improper. Emple 6 Evlute Figure () d (b) 4 d ( ) /3 Solution (). The integrl is improper becuse the integrnd pproches s pproches the lower limit from the right (Figure 7.8.). From Definition we obtin d = lim k + k d = lim k + [ ] ln k = lim k + [ ] ln +ln k = lim ln k = k + = so the integrl diverges. Solution (b). The integrl is improper becuse the integrnd pproches + t =, which is inside the intervl of integrtion. From Definition we obtin Figure d ( ) /3 = d ( ) /3 + 4 d ( ) /3 (7) nd we must investigte the convergence of both improper integrls on the right. Since 4 d = lim ( ) /3 k d = lim ( ) /3 k + k 4 k d [ = lim 3(k ) /3 3( ) /3] = 3 ( ) /3 k d [ = lim 3(4 ) /3 3(k ) /3] = 3 3 ( ) /3 k + we hve from (7) tht 4 d ( ) /3 =

141 7.8 Improper Integrls 553 WARNING It is sometimes tempting to ppl the Fundmentl Theorem of Clculus directl to n improper integrl without tking the pproprite limits. To illustrte wht cn go wrong with this procedure, suppose we fil to recognize tht the integrl d (8) ( ) is improper nd mistkenl evlute this integrl s ] = () = This result is clerl incorrect becuse the integrnd is never negtive nd hence the integrl cnnot be negtive! To evlute (8) correctl we should first write d ( ) = d ( ) + nd then tret ech term s n improper integrl. For the first term, so (8) diverges. d ( ) = lim k k d ( ) = lim k d ( ) [ k ] =+ ARC LENGTH AND SURFACE AREA USING IMPROPER INTEGRALS In Definitions 6.4. nd 6.5. for rc length nd surfce re we required the function f to be smooth (continuous first derivtive) to ensure the integrbilit in the resulting formul. However, smoothness is overl restrictive since some of the most bsic formuls in geometr involve functions tht re not smooth but led to convergent improper integrls. Accordingl, let us gree to etend the definitions of rc length nd surfce re to llow functions tht re not smooth, but for which the resulting integrl in the formul converges. Emple 7 Derive the formul for the circumference of circle of rdius r. r Figure 7.8. = r r Solution. For convenience, let us ssume tht the circle is centered t the origin, in which cse its eqution is + = r. We will find the rc length of the portion of the circle tht lies in the first qudrnt nd then multipl b 4 to obtin the totl circumference (Figure 7.8.). Since the eqution of the upper semicircle is = r, it follows from Formul (4) of Section 6.4 tht the circumference C is r r ( C = 4 + (d/d) d = 4 + ) r d d = 4r r This integrl is improper becuse of the infinite discontinuit t = r, nd hence we evlute it b writing r C = 4r lim k r k [ k r = 4r lim = 4r lim k r d r ( sin )] k r [ ( k sin r ) sin Formul (77) in the Endpper Integrl Tble ( π ) = 4r[sin sin ]=4r = πr ]

142 554 Chpter 7 / Principles of Integrl Evlution QUICK CHECK EXERCISES 7.8 (See pge 557 for nswers.). In ech prt, determine whether the integrl is improper, nd if so, eplin wh. Do not evlute the integrls. () (c) 3π/4 π/4 + cot d + d (b) (d) π π/4 + cot d d. Epress ech improper integrl in Quick Check Eercise in terms of one or more pproprite limits. Do not evlute the limits. 3. The improper integrl + p d converges to provided. 4. Evlute the integrls tht converge. () (c) + e d 3 d (b) (d) + e d 3 d EXERCISE SET 7.8 Grphing Utilit C CAS. In ech prt, determine whether the integrl is improper, nd if so, eplin wh. 5 d 5 d () (b) (c) ln d (d) e d π/4 d (e) 3 (f ) tn d. In ech prt, determine ll vlues of p for which the integrl is improper. d d () (b) (c) e p d p p 3 3 Evlute the integrls tht converge e d 4. + d d 6. e d e ln 3 d 8. ln d d 3 d 9.. ( ) π/ e 3 d. d 4. d 6. ( + 3) d 8. ( 4) tn d. d e d 3 e d + e t dt + e t d 3 d 4 d π/ π/ sin cos d 4. d 6. /3 d 8. d 3. d ( + ) 3. π/4 + + sec tn d d d ( ) /3 d d ( + ) True Flse Determine whether the sttement is true or flse. Eplin our nswer /3 d converges to If f is continuous on [,+ ) nd lim + f() =, then + f()d converges d is n improper integrl. ( 3) 3 d = 37 4 Mke the u-substitution nd evlute the resulting definite integrl. + e 37. d; u = [Note: u + s +.] d ( + 4) ; u = e d; u = e e [Note: u s +.] [Note: u + s +.]

143 C C C 4. + e d; u = e e 4 4 Epress the improper integrl s limit, nd then evlute tht limit with CAS. Confirm the nswer b evluting the integrl directl with the CAS e cos d 4. e 3 d In ech prt, tr to evlute the integrl ectl with CAS. If our result is not simple numericl nswer, then use the CAS to find numericl pproimtion of the integrl. () d (b) + + ln sin (c) d (d) e In ech prt, confirm the result with CAS. + sin π + () d = (b) (c) ln d = π d d e d = π 45. Find the length of the curve = (4 /3 ) 3/ over the intervl [, 8]. 46. Find the length of the curve = 4 over the intervl [, ] Use L Hôpitl s rule to help evlute the improper integrl. + ln 47. ln d 48. d 49. Find the re of the region between the -is nd the curve = e 3 for. 5. Find the re of the region between the -is nd the curve = 8/( 4) for Suppose tht the region between the -is nd the curve = e for is revolved bout the -is. () Find the volume of the solid tht is generted. (b) Find the surfce re of the solid. 7.8 Improper Integrls Use the results in Eercise () Confirm grphicll nd lgebricll tht e e ( ) (b) Evlute the integrl + e d (c) Wht does the result obtined in prt (b) tell ou bout the integrl + e d? 54. () Confirm grphicll nd lgebricll tht + e ( ) + (b) Evlute the integrl + d + (c) Wht does the result obtined in prt (b) tell ou bout the integrl + e + d? 55. Let R be the region to the right of = tht is bounded b the -is nd the curve = /. When this region is revolved bout the -is it genertes solid whose surfce is known s Gbriel s Horn (for resons tht should be cler from the ccompning figure). Show tht the solid hs finite volume but its surfce hs n infinite re. [Note: It hs been suggested tht if one could sturte the interior of the solid with pint nd llow it to seep through to the surfce, then one could pint n infinite surfce with finite mount of pint! Wht do ou think?] = FOCUS ON CONCEPTS 5. Suppose tht f nd g re continuous functions nd tht f() g() if. Give resonble informl rgument using res to eplin wh the following results re true. () If + f()d diverges, then + g() d diverges. (b) If + g() d converges, then + f()d converges nd + f()d + g() d. [Note: The results in this eercise re sometimes clled comprison tests for improper integrls.] Figure E In ech prt, use Eercise 5 to determine whether the integrl converges or diverges. If it converges, then use prt (b) of tht eercise to find n upper bound on the vlue of the integrl. () (c) d e + d (b) d

144 556 Chpter 7 / Principles of Integrl Evlution C FOCUS ON CONCEPTS 57. Sketch the region whose re is + d + nd use our sketch to show tht + d + = d 58. () Give resonble informl rgument, bsed on res, tht eplins wh the integrls diverge. + (b) Show tht sin d + cos nd + d diverges. cos d 59. In electromgnetic theor, the mgnetic potentil t point on the is of circulr coil is given b u = πni r + d k (r + ) 3/ where N,I,r,k, nd re constnts. Find u. 6. The verge speed, v, of the molecules of n idel gs is given b v = 4 ( ) M 3/ + v 3 e Mv /(RT ) dv π RT nd the root-men-squre speed, v rms,b vrms = 4 ( ) M 3/ + v 4 e Mv /(RT ) dv π RT where v is the moleculr speed, T is the gs temperture, M is the moleculr weight of the gs, nd R is the gs constnt. () Use CAS to show tht + 3 e d =, > 4 nd use this result to show tht v = 8RT /(πm). (b) Use CAS to show tht + 4 e d = 3 π 8, > 5 nd use this result to show tht v rms = 3RT /M. 6 6 Mediction cn be dministered to ptient using vriet of methods. For given method, let c(t) denote the concentrtion of mediction in the ptient s bloodstrem (mesured in mg/l) t hours fter the dose is given. The re under the curve c = c(t) over the time intervl [, + ) indictes the vilbilit of the mediction for the ptient s bod. Determine which method provides the greter vilbilit. 6. Method : c (t) = 5(e.t e t ); Method : c (t) = 4(e.t e 3t ) 6. Method : c (t) = 6(e.4t e.3t ); Method : c (t) = 5(e.4t e 3t ) C 63. In Eercise 5 of Section 6.6, we determined the work required to lift 6 lb stellite to n orbitl position tht is mi bove the Erth s surfce. The ides discussed in tht eercise will be needed here. () Find definite integrl tht represents the work required to lift 6 lb stellite to position b miles bove the Erth s surfce. (b) Find definite integrl tht represents the work required to lift 6 lb stellite n infinite distnce bove the Erth s surfce. Evlute the integrl. [Note: The result obtined here is sometimes clled the work required to escpe the Erth s grvit.] A trnsform is formul tht converts or trnsforms one function into nother. Trnsforms re used in pplictions to convert difficult problem into n esier problem whose solution cn then be used to solve the originl difficult problem. The Lplce trnsform of function f(t),which pls n importnt role in the stud of differentil equtions, is denoted b {f(t)} nd is defined b {f(t)}= + e st f(t)dt In this formul s is treted s constnt in the integrtion process; thus, the Lplce trnsform hs the effect of trnsforming f(t)into function of s. Use this formul in these eercises. 64. Show tht () {} = s, s> (b) {et }= s, s> (c) {sin t} = s +, s> (d) {cos t} = s s +, s>. 65. In ech prt, find the Lplce trnsform. () f(t) = t, { s> (b) f(t) = t, s>, t < 3 (c) f(t) =, t 3, s > 66. Lter in the tet, we will show tht + e d = π Confirm tht this is resonble b using CAS or clcultor with numericl integrtion cpbilit. 67. Use the result in Eercise 66 to show tht + π () e d =, > + (b) e /σ d =, σ>. πσ A convergent improper integrl over n infinite intervl cn be pproimted b first replcing the infinite limit(s) of integrtion b finite limit(s), then using numericl integrtion technique, such s Simpson s rule, to pproimte the integrl with finite limit(s). This technique is illustrted in these eercises.

145 68. Suppose tht the integrl in Eercise 66 is pproimted b first writing it s + e d = K e d + + K e d then dropping the second term, nd then ppling Simpson s rule to the integrl K e d The resulting pproimtion hs two sources of error: the error from Simpson s rule nd the error E = + K e d tht results from discrding the second term. We cll E the trunction error. () Approimte the integrl in Eercise 66 b ppling Simpson s rule with n = subdivisions to the integrl 3 e d Round our nswer to four deciml plces nd compre it to π rounded to four deciml plces. (b) Use the result tht ou obtined in Eercise 5 nd the fct tht e 3 e for 3 to show tht the trunction error for the pproimtion in prt () stisfies <E< () It cn be shown tht d = π 3 Approimte this integrl b ppling Simpson s rule with n = subdivisions to the integrl d Round our nswer to three deciml plces nd compre it to π/3 rounded to three deciml plces. QUICK CHECK ANSWERS 7.8 C Chpter 7 Review Eercises 557 (b) Use the result tht ou obtined in Eercise 5 nd the fct tht /( 6 + ) </ 6 for 4 to show tht the trunction error for the pproimtion in prt () stisfies <E< For wht vlues of p does e p d converge? 7. Show tht d/ p converges if p< nd diverges if p. 7. It is sometimes possible to convert n improper integrl into proper integrl hving the sme vlue b mking n pproprite substitution. Evlute the following integrl b mking the indicted substitution, nd investigte wht hppens if ou evlute the integrl directl using CAS. + d; u = Trnsform the given improper integrl into proper integrl b mking the stted u-substitution; then pproimte the proper integrl b Simpson s rule with n = subdivisions. Round our nswer to three deciml plces. cos 73. d; u = 74. sin d; u = 75. Writing Wht is improper bout n integrl over n infinite intervl? Eplin wh Definition 5.5. for b f()d fils for + f()d. Discuss strteg for ssigning vlue to + f()d. 76. Writing Wht is improper bout definite integrl over n intervl on which the integrnd hs n infinite discontinuit? Eplin wh Definition 5.5. for b f()d fils if the grph of f hs verticl smptote t =. Discuss strteg for ssigning vlue to b f()d in this circumstnce.. () proper (b) improper, since cot hs n infinite discontinuit t = π (c) improper, since there is n infinite intervl of integrtion (d) improper, since there is n infinite intervl of integrtion nd the integrnd hs n infinite discontinuit t = b b. (b) lim cot d (c) lim d (d) b π π/4 b + + lim b d + lim + b + d 3. p ; p> 4. () (b) diverges (c) diverges (d) 3 CHAPTER 7 REVIEW EXERCISES 6 Evlute the given integrl with the id of n pproprite u-substitution d. sec π d cos d 3. sin d 4. ln 9 5. tn ( ) sec ( )d d 7. () Evlute the integrl d three ws: using the substitution u =, using the substitution u =, nd completing the squre. (b) Show tht the nswers in prt () re equivlent.

146 558 Chpter 7 / Principles of Integrl Evlution 3 8. Evlute the integrl + d () using integrtion b prts (b) using the substitution u = +. 9 Use integrtion b prts to evlute the integrl. 9. e d. sin d. ln( + 3)d. / tn ()d 3. Evlute 8 4 cos dusing tbulr integrtion b prts. 4. A prticle moving long the -is hs velocit function v(t) = t e t. How fr does the prticle trvel from time t = tot = 5? 5 Evlute the integrl. 5. sin 5θdθ 6. sin 3 cos d sin cos d 8. sin 4 d. π/6 sin cos 4d cos 5 ( )d 6 Evlute the integrl b mking n pproprite trigonometric substitution.. d. d 9 6 d d d Evlute the integrl using the method of prtil frctions. d d d 3. + ( )( 3) d d 3. d 3. ( + ) Consider the integrl 3 d. () Evlute the integrl using the substitution = sec θ. For wht vlues of is our result vlid? (b) Evlute the integrl using the substitution = sin θ. For wht vlues of is our result vlid? (c) Evlute the integrl using the method of prtil frctions. For wht vlues of is our result vlid? 34. Find the re of the region tht is enclosed b the curves = ( 3)/( 3 + ), =, =, nd =. d 35 4 Use the Endpper Integrl Tble to evlute the integrl. 35. sin 7 cos 9d 36. ( 3 )e d 37. d d tn d 4. + d 4 4 Approimte the integrl using () the midpoint pproimtion M, (b) the trpezoidl pproimtion T, nd (c) Simpson s rule pproimtion S. In ech cse, find the ect vlue of the integrl nd pproimte the bsolute error. Epress our nswers to t lest four deciml plces d 4. + d Use inequlities (), (3), nd (4) of Section 7.7 to find upper bounds on the errors in prts (), (b), or (c) of the indicted eercise. 43. Eercise Eercise Use inequlities (), (3), nd (4) of Section 7.7 to find number n of subintervls for () the midpoint pproimtion M n, (b) the trpezoidl pproimtion T n, nd (c) Simpson s rule pproimtion S n to ensure the bsolute error will be less thn Eercise Eercise Evlute the integrl if it converges e d d d 9 5. d 5. Find the re tht is enclosed between the -is nd the curve = (ln )/ for e. 5. Find the volume of the solid tht is generted when the region between the -is nd the curve = e for is revolved bout the -is. 53. Find positive vlue of tht stisfies the eqution + + d = 54. Consider the following methods for evluting integrls: u-substitution, integrtion b prts, prtil frctions, reduction formuls, nd trigonometric substitutions. In ech prt, stte the pproch tht ou would tr first to evlute the integrl. If none of them seems pproprite, then s so. You need not evlute the integrl. () sin d (b) cos sin d (cont.)

147 (c) (e) (g) (i) tn 7 d (d) tn 7 sec d 3 3 d (f ) 3 + ( + ) d 3 4 tn d (h) d 4 d Evlute the integrl. d cos 3d (3 + ) 3/ π/4 57. tn 7 cos θ θdθ 58. sin θ 6 sin θ + dθ sin cos 3 d 6. ( 3) d / 6. e cos 3d 6. ( ) 3/ d / Chpter 7 Mking Connections 559 /3 d 64. ( )( + )( 3) 8 4 d e + d 68. / + + sin d d ( + ) d d + b,,b > ln 7. d (4 9 ) e d d ( + + ) tn 5 4 sec 4 4d sec θ tn 3 θ tn θ dθ CHAPTER 7 MAKING CONNECTIONS C CAS. Recll from Theorem 3.3. nd the discussion preceding it tht if f () >, then the function f is incresing nd hs n inverse function. Prts (), (b), nd (c) of this problem show tht if this condition is stisfied nd if f is continuous, then definite integrl of f cn be epressed in terms of definite integrl of f. () Use integrtion b prts to show tht b f()d = bf(b) f() b f () d (b) Use the result in prt () to show tht if = f(), then b f()d = bf(b) f() f(b) f() f () d (c) Show tht if we let α = f() nd β = f(b), then the result in prt (b) cn be written s β α f () d = βf (β) αf (α) f (β) f (α) f()d. In ech prt, use the result in Eercise to obtin the eqution, nd then confirm tht the eqution is correct b performing the integrtions. () (b) / e e sin d= sin ( ln d= (e e) ) π/6 sin d e d C 3. The Gmm function, Ɣ(), is defined s Ɣ() = + t e t dt It cn be shown tht this improper integrl converges if nd onl if >. () Find Ɣ(). (b) Prove: Ɣ( + ) = Ɣ() for ll >. [Hint: Use integrtion b prts.] (c) Use the results in prts () nd (b) to find Ɣ(), Ɣ(3), nd Ɣ(4); nd then mke conjecture bout Ɣ(n) for positive integer vlues of n. (d) Show tht Ɣ ( ) = π.[hint: See Eercise 66 of Section 7.8.] (e) Use the results obtined in prts (b) nd (d) to show tht Ɣ ( ) 3 = ( π nd Ɣ 5 ) = 3 4 π. 4. Refer to the Gmm function defined in Eercise 3 to show tht () (ln ) n d = ( ) n Ɣ(n + ), n> [Hint: Let t = ln.] + ( ) n + (b) e n d = Ɣ, n>. n [Hint: Let t = n. Use the result in Eercise 3(b).] 5. A simple pendulum consists of mss tht swings in verticl plne t the end of mssless rod of length L, s shown in the ccompning figure. Suppose tht simple pendulum is displced through n ngle θ nd relesed from rest. It cn be

148 56 Chpter 7 / Principles of Integrl Evlution shown tht in the bsence of friction, the time T required for the pendulum to mke one complete bck-nd-forth swing, clled the period,isgivenb T = 8L g θ cos θ cos θ dθ () where θ = θ(t) is the ngle the pendulum mkes with the verticl t time t. The improper integrl in () is difficult to evlute numericll. B substitution outlined below it cn be shown tht the period cn be epressed s T = 4 L g π/ dφ () k sin φ where k = sin(θ /). The integrl in () is clled complete elliptic integrl of the first kind nd is more esil evluted b numericl methods. () Obtin () from () b substituting cos θ = sin (θ/) cos θ = sin (θ /) k = sin(θ /) nd then mking the chnge of vrible sin φ = sin(θ /) sin(θ /) = sin(θ /) k (b) Use () nd the numericl integrtion cpbilit of our CAS to estimte the period of simple pendulum for which L =.5 ft, θ =, nd g = 3 ft/s. u L Figure E-5 E XPANDING THE C ALCULUS H ORIZON To lern how numericl integrtion cn be pplied to the cost nlsis of n engineering project, see the module entitled Rilrod Design t:

149 8 Photo b Milton Bell, ctlog # Lu-mb, Tes Archeologicl Reserch Lbrtor, The Universit of Tes t Austin MATHEMATICAL MODELING WITH DIFFERENTIAL EQUATIONS In the 9s, ecvtion of n rcheologicl site in Folsom, New Meico, uncovered collection of prehistoric stone sperheds now known s Folsom points. In 95, crbon dting of chrred bison bones found nerb confirmed tht humn hunters lived in the re between 9 B.C. nd 8 B.C. We will stud crbon dting in this chpter. Mn fundmentl lws of science nd engineering cn be epressed in terms of differentil equtions. We introduced the concept of differentil eqution in Section 5., but in this chpter we will go into more detil. We will discuss some importnt mthemticl models tht involve differentil equtions, nd we will discuss some methods for solving nd pproimting solutions of some of the bsic tpes of differentil equtions. However, we will onl be ble to touch the surfce of this topic, leving mn importnt topics in differentil equtions to courses tht re devoted completel to the subject. 8. MODELING WITH DIFFERENTIAL EQUATIONS In this section we will introduce some bsic terminolog nd concepts concerning differentil equtions. We will lso discuss the generl ide of modeling with differentil equtions, nd we will encounter importnt models tht cn be pplied to demogrph, medicine, ecolog, nd phsics. In lter sections of this chpter we will investigte methods tht m be used to solve these differentil equtions. Tble 8.. differentil eqution d d = 3 d d d = d d 3 d dt 3 t + (t ) = e t dt = e + = cos t order 3 TERMINOLOGY Recll from Section 5. tht differentil eqution is n eqution involving one or more derivtives of n unknown function. In this section we will denote the unknown function b = () unless the differentil eqution rises from n pplied problem involving time, in which cse we will denote it b = (t). The order of differentil eqution is the order of the highest derivtive tht it contins. Some emples re given in Tble 8... The lst two equtions in tht tble re epressed in prime nottion, which does not specif the independent vrible eplicitl. However, ou will usull be ble to tell from the eqution itself or from the contet in which it rises whether to interpret s d/d or d/dt. SOLUTIONS OF DIFFERENTIAL EQUATIONS A function = () is solution of differentil eqution on n open intervl if the eqution is stisfied identicll on the intervl when nd its derivtives re substituted 56

150 56 Chpter 8 / Mthemticl Modeling with Differentil Equtions The first-order eqution () hs single rbitrr constnt in its generl solution (). Usull, the generl solution of n nth-order differentil eqution will contin n rbitrr constnts. This is plusible, since n integrtions re needed to recover function from its nth derivtive. into the eqution. For emple, = e is solution of the differentil eqution d d = e () on the intervl (, + ), since substituting nd its derivtive into the left side of this eqution ields d d = d d [e ] e = e e = e for ll rel vlues of. However, this is not the onl solution on (, + ); for emple, the function = e + Ce () is lso solution for ever rel vlue of the constnt C, since d d = d d [e + Ce ] (e + Ce ) = (e + Ce ) (e + Ce ) = e After developing some techniques for solving equtions such s (), we will be ble to show tht ll solutions of () on (, + ) cn be obtined b substituting vlues for the constnt C in (). On given intervl, solution of differentil eqution from which ll solutions on tht intervl cn be derived b substituting vlues for rbitrr constnts is clled generl solution of the eqution on the intervl. Thus () is generl solution of () on the intervl (, + ). The grph of solution of differentil eqution is clled n integrl curve for the eqution, so the generl solution of differentil eqution produces fmil of integrl curves corresponding to the different possible choices for the rbitrr constnts. For emple, Figure 8.. shows some integrl curves for (), which were obtined b ssigning vlues to the rbitrr constnt in (). = e + Ce C = 3 C = C = 4 C = C = C = C = Figure 8.. Integrl curves for d d = e INITIAL-VALUE PROBLEMS When n pplied problem leds to differentil eqution, there re usull conditions in the problem tht determine specific vlues for the rbitrr constnts. As rule of thumb, it requires n conditions to determine vlues for ll n rbitrr constnts in the generl solution of n nth-order differentil eqution (one condition for ech constnt). For firstorder eqution, the single rbitrr constnt cn be determined b specifing the vlue of the unknown function () t n rbitrr -vlue, s ( ) =. This is clled n initil condition, nd the problem of solving first-order eqution subject to n initil condition is clled first-order initil-vlue problem. Geometricll, the initil condition ( ) = hs the effect of isolting the integrl curve tht psses through the point (, ) from the complete fmil of integrl curves.

151 8. Modeling with Differentil Equtions 563 Emple The solution of the initil-vlue problem d d = e, () = 3 cn be obtined b substituting the initil condition =, = 3 in the generl solution () to find C. We obtin 3 = e + Ce = + C Thus, C =, nd the solution of the initil-vlue problem, which is obtined b substituting this vlue of C in (), is = e + e Geometricll, this solution is relized s the integrl curve in Figure 8.. tht psses through the point (, 3). Since mn importnt principles in the phsicl nd socil sciences involve rtes of chnge, it should not be surprising tht such principles cn often be modeled b differentil equtions. Here re some emples of the modeling process. AndresReh/iStockphoto When the number of bcteri is smll, n uninhibited popultion growth model cn be used to model the growth of bcteri in petri dish. UNINHIBITED POPULATION GROWTH One of the simplest models of popultion growth is bsed on the observtion tht when popultions (people, plnts, bcteri, nd fruit flies, for emple) re not constrined b environmentl limittions, the tend to grow t rte tht is proportionl to the size of the popultion the lrger the popultion, the more rpidl it grows. To trnslte this principle into mthemticl model, suppose tht = (t) denotes the popultion t time t. At ech point in time, the rte of increse of the popultion with respect to time is d/dt, so the ssumption tht the rte of growth is proportionl to the popultion is described b the differentil eqution d = k (3) dt where k is positive constnt of proportionlit tht cn usull be determined eperimentll. Thus, if the popultion is known t some point in time, s = t time t =, then formul for the popultion (t) cn be obtined b solving the initil-vlue problem d dt = k, () = INHIBITED POPULATION GROWTH; LOGISTIC MODELS The uninhibited popultion growth model ws predicted on the ssumption tht the popultion = (t) ws not constrined b the environment. While this ssumption is resonble s long s the size of the popultion is reltivel smll, environmentl effects become incresingl importnt s the popultion grows. In generl, popultions grow within ecologicl sstems tht cn onl support certin number of individuls; the number L of such individuls is clled the crring cpcit of the sstem. When >L, the popultion eceeds the cpcit of the ecologicl sstem nd tends to decrese towrd L; when <L, the popultion is below the cpcit of the ecologicl sstem nd tends to increse towrd L; when = L, the popultion is in blnce with the cpcit of the ecologicl sstem nd tends to remin stble. To trnslte this into mthemticl model, we must look for differentil eqution in which >, L>, nd d dt < if L >, d dt > if L <, d dt = if L =

152 564 Chpter 8 / Mthemticl Modeling with Differentil Equtions Moreover, when the popultion is fr below the crring cpcit (i.e., /L ), then the environmentl constrints should hve little effect, nd the growth rte should behve like the uninhibited popultion model. Thus, we wnt d k if dt L A simple differentil eqution tht meets ll of these requirements is d ( = k ) dt L where k is positive constnt of proportionlit. Thus if k nd L cn be determined eperimentll, nd if the popultion is known t some point, s () =, then formul for the popultion (t) cn be determined b solving the initil-vlue problem d ( = k ), () = (4) dt L This theor of popultion growth is due to the Belgin mthemticin P. F. Verhulst (84 849), who introduced it in 838 nd described it s logistic growth. Thus, the differentil eqution in (4) is clled the logistic differentil eqution, nd the growth model described b (4) is clled the logistic model. PHARMACOLOGY When drug (s, penicillin or spirin) is dministered to n individul, it enters the bloodstrem nd then is bsorbed b the bod over time. Medicl reserch hs shown tht the mount of drug tht is present in the bloodstrem tends to decrese t rte tht is proportionl to the mount of the drug present the more of the drug tht is present in the bloodstrem, the more rpidl it is bsorbed b the bod. To trnslte this principle into mthemticl model, suppose tht = (t)is the mount of the drug present in the bloodstrem t time t. At ech point in time, the rte of chnge in with respect to t is d/dt, so the ssumption tht the rte of decrese is proportionl to the mount in the bloodstrem trnsltes into the differentil eqution d = k (5) dt where k is positive constnt of proportionlit tht depends on the drug nd cn be determined eperimentll. The negtive sign is required becuse decreses with time. Thus, if the initil dosge of the drug is known, s = t time t =, then formul for (t) cn be obtined b solving the initil-vlue problem d = k, () = dt SPREAD OF DISEASE Suppose tht disese begins to spred in popultion of L individuls. Logic suggests tht t ech point in time the rte t which the disese spreds will depend on how mn individuls re lred ffected nd how mn re not s more individuls re ffected, the opportunit to spred the disese tends to increse, but t the sme time there re fewer individuls who re not ffected, so the opportunit to spred the disese tends to decrese. Thus, there re two conflicting influences on the rte t which the disese spreds. Verhulst s model fell into obscurit for nerl hundred ers becuse he did not hve sufficient census dt to test its vlidit. However, interest in the model ws revived during the 93s when biologists used it successfull to describe the growth of fruit fl nd flour beetle popultions. Verhulst himself used the model to predict tht n upper limit of Belgium s popultion would be pproimtel 9,4,. In 6 the popultion ws bout,379,.

153 8. Modeling with Differentil Equtions 565 Show tht the model for the spred of disese cn be viewed s logistic model with constnt of proportionlit kl b rewriting (6) ppropritel. Nturl position m To trnslte this into mthemticl model, suppose tht = (t) is the number of individuls who hve the disese t time t, so of necessit the number of individuls who do not hve the disese t time t is L. As the vlue of increses, the vlue of L decreses, so the conflicting influences of the two fctors on the rte of spred d/dt re tken into ccount b the differentil eqution d dt = k(l ) where k is positive constnt of proportionlit tht depends on the nture of the disese nd the behvior ptterns of the individuls nd cn be determined eperimentll. Thus, if the number of ffected individuls is known t some point in time, s = t time t =, then formul for (t) cn be obtined b solving the initil-vlue problem d dt = k(l ), () = (6) NEWTON S LAW OF COOLING If hot object is plced into cool environment, the object will cool t rte proportionl to the difference in temperture between the object nd the environment. Similrl, if cold object is plced into wrm environment, the object will wrm t rte tht is gin proportionl to the difference in temperture between the object nd the environment. Together, these observtions comprise result known s Newton s Lw of Cooling. (Newton s Lw of Cooling ppered previousl in the eercises of Section. nd ws mentioned briefl in Section 5.8.) To trnslte this into mthemticl model, suppose tht T = T(t)is the temperture of the object t time t nd tht T e is the temperture of the environment, which is ssumed to be constnt. Since the rte of chnge dt/dt is proportionl to T T e,we hve dt = k(t T e ) dt where k is constnt of proportionlit. Moreover, since dt/dt is positive when T<T e, nd is negtive when T>T e, the sign of k must be negtive. Thus if the temperture of the object is known t some time, s T = T t time t =, then formul for the temperture T(t)cn be obtined b solving the initil-vlue problem dt dt = k(t T e ), T () = T (7) m Figure 8.. Figure 8..3 Stretched m Relesed Nturl position m VIBRATIONS OF SPRINGS We conclude this section with n engineering model tht leds to second-order differentil eqution. As shown in Figure 8.., consider block of mss m ttched to the end of horizontl spring. Assume tht the block is then set into vibrtor motion b pulling the spring beond its nturl position nd relesing it t time t =. We will be interested in finding mthemticl model tht describes the vibrtor motion of the block over time. To trnslte this problem into mthemticl form, we introduce horizontl -is whose positive direction is to the right nd whose origin is t the right end of the spring when the spring is in its nturl position (Figure 8..3). Our gol is to find model for the coordinte = (t) of the point of ttchment of the block to the spring s function of time. In developing this model, we will ssume tht the onl force on the mss m is the restoring force of the spring, nd we will ignore the influence of other forces such s friction, ir resistnce, nd so forth. Recll from Hooke s Lw (Section 6.6) tht when the connection point hs coordinte (t), the restoring force is k(t), where k is the spring constnt. [The negtive sign is due to the fct tht the restoring force is to the left when (t) is positive, nd the restoring force is to the right when (t) is negtive.] It follows from Newton s

154 566 Chpter 8 / Mthemticl Modeling with Differentil Equtions Second Lw of Motion [Eqution (5) of Section 6.6] tht this restoring force is equl to the product of the mss m nd the ccelertion d /dt of the mss. In other words, we hve m d dt = k which is second-order differentil eqution for. If t time t = the mss is relesed from rest t position () =, then formul for (t) cn be found b solving the initil-vlue problem m d dt = k, () =, () = (8) [If t time t = the mss is given n initil velocit v =, then the condition () = must be replced b () = v.] QUICK CHECK EXERCISES 8. (See pge 568 for nswers.). Mtch ech differentil eqution with its fmil of solutions. () d d = (i) = + C (b) = 4 (ii) = C sin + C cos (c) d d = (iii) = C e + C e (d) d d = 4 (iv) = C. If = C e + C e is the generl solution of differentil eqution, then the order of the eqution is, nd solution to the differentil eqution tht stisfies the initil conditions () =, () = 4 is given b =. 3. The grph of differentible function = () psses through the point (, ) nd t ever point P(,) on the grph the tngent line is perpendiculr to the line through P nd the origin. Find n initil-vlue problem whose solution is (). 4. A glss of ice wter with temperture of 36 F is plced in room with constnt temperture of 68 F. Assuming tht Newton s Lw of Cooling pplies, find n initil-vlue problem whose solution is the temperture of wter t minutes fter it is plced in the room. [Note: The differentil eqution will involve constnt of proportionlit.] EXERCISE SET 8.. Confirm tht = 3e 3 is solution of the initil-vlue problem = 3,() = 3.. Confirm tht = cos + is solution of the initil-vlue problem = 3 sin,() = Stte the order of the differentil eqution, nd confirm tht the functions in the given fmil re solutions. 3. () ( + ) d = ; = c( + ) d (b) + = ; = c sin t + c cos t 4. () d d + = ; = ce / + 3 (b) = ; = c e t + c e t 5 8 True Flse Determine whether the sttement is true or flse. Eplin our nswer. 5. The eqution ( ) d = d d d + is n emple of second-order differentil eqution. 6. The differentil eqution d d = + hs solution tht is constnt. 7. We epect the generl solution of the differentil eqution d 3 d + 3 d 3 d d d + 4 = to involve three rbitrr constnts. 8. If ever solution to differentil eqution cn be epressed in the form = Ae +b for some choice of constnts A nd b, then the differentil eqution must be of second order. 9 4 In ech prt, verif tht the functions re solutions of the differentil eqution b substituting the functions into the eqution = () e nd e (b) c e + c e (c, c constnts)

155 8. Modeling with Differentil Equtions = () e nd e 3 (b) c e + c e = () e nd e (b) c e + c e (c, c constnts) (c, c constnts) = () e 4 nd e 4 (b) c e 4 + c e 4 (c, c constnts) = () sin nd cos (b) c sin + c cos (c, c constnts) = () e sin 3 nd e cos 3 (b) e (c sin 3 + c cos 3) (c, c constnts) 5 Use the results of Eercises 9 4 to find solution to the initil-vlue problem =, () =, () = =, () =, () = =, () =, () = =, () =, () = =, () =, () = =, () =, () = 6 Find solution to the initil-vlue problem =, () = =, () =, () = 3. =, () = [Hint: Assume the solution hs n inverse function = (). Find, nd solve, differentil eqution tht involves ().] 4. = +, () = (See Eercise 3.) 5. + =, () = [Hint: Interpret the left-hnd side of the eqution s the derivtive of product of two functions.] 6. + = e, () = + e (See Eercise 5.) FOCUS ON CONCEPTS 7. () Suppose tht quntit = (t) increses t rte tht is proportionl to the squre of the mount present, nd suppose tht t time t =, the mount present is. Find n initil-vlue problem whose solution is (t). (b) Suppose tht quntit = (t) decreses t rte tht is proportionl to the squre of the mount present, nd suppose tht t time t =, the mount present is. Find n initil-vlue problem whose solution is (t). 8. () Suppose tht quntit = (t) chnges in such w tht d/dt = k, where k>. Describe how chnges in words. (b) Suppose tht quntit = (t) chnges in such w tht d/dt = k 3, where k>. Describe how chnges in words. 9. () Suppose tht prticle moves long n s-is in such w tht its velocit v(t) is lws hlf of s(t). Find differentil eqution whose solution is s(t). (b) Suppose tht n object moves long n s-is in such w tht its ccelertion (t) is lws twice the velocit. Find differentil eqution whose solution is s(t). 3. Suppose tht bod moves long n s-is through resistive medium in such w tht the velocit v = v(t) decreses t rte tht is twice the squre of the velocit. () Find differentil eqution whose solution is the velocit v(t). (b) Find differentil eqution whose solution is the position s(t). 3. Consider solution = (t) to the uninhibited popultion growth model. () Use Eqution (3) to eplin wh will be n incresing function of t. (b) Use Eqution (3) to eplin wh the grph = (t) will be concve up. 3. Consider the logistic model for popultion growth. () Eplin wh there re two constnt solutions to this model. (b) For wht size of the popultion will the popultion be growing most rpidl? 33. Consider the model for the spred of disese. () Eplin wh there re two constnt solutions to this model. (b) For wht size of the infected popultion is the disese spreding most rpidl? 34. Eplin wh there is ectl one constnt solution to the Newton s Lw of Cooling model. 35. Show tht if c nd c re n constnts, the function ( ) ( ) k k = (t) = c cos m t + c sin m t is solution to the differentil eqution for the vibrting spring. (The corresponding motion of the spring is referred to s simple hrmonic motion.) 36. () Use the result of Eercise 35 to solve the initil-vlue problem in (8). (b) Find the mplitude, period, nd frequenc of our nswer to prt (), nd interpret ech of these in terms of the motion of the spring. 37. Writing Select one of the models in this section nd write prgrph tht discusses conditions under which the model would not be pproprite. How might ou modif the model to tke those conditions into ccount?

156 568 Chpter 8 / Mthemticl Modeling with Differentil Equtions QUICK CHECK ANSWERS 8.. () (iv) (b) (iii) (c) (i) (d) (ii). ; e + e 3. d d =, () = 4. dt dt = k(t 68), T() = SEPARATION OF VARIABLES In this section we will discuss method, clled seprtion of vribles, tht cn be used to solve lrge clss of first-order differentil equtions of prticulr form. We will use this method to investigte mthemticl models for eponentil growth nd dec, including popultion models nd crbon dting. Some writers define seprble eqution to be one tht cn be written in the form d/d = G()H (). Eplin wh this is equivlent to our definition. FIRST-ORDER SEPARABLE EQUATIONS We will now consider method of solution tht cn often be pplied to first-order equtions tht re epressible in the form h() d = g() () d Such first-order equtions re sid to be seprble. Some emples of seprble equtions re given in Tble 8... The nme seprble rises from the fct tht Eqution () cn be rewritten in the differentil form h() d = g() d () in which the epressions involving nd pper on opposite sides. rewriting () in form () is clled seprting vribles. The process of Tble 8.. eqution form () h() g() d d = d d = 3 d d = d d = d = d d 3 = d 3 d = d d = d To motivte method for solving seprble equtions, ssume tht h() nd g() re continuous functions of their respective vribles, nd let H()nd G() denote ntiderivtives of h() nd g(), respectivel. Consider the eqution tht results if we integrte both sides of (), the left side with respect to nd the right side with respect to. We then hve h() d = g() d (3) or, equivlentl, H() = G() + C (4) where C denotes constnt. We clim tht differentible function = () is solution to () if nd onl if stisfies Eqution (4) for some choice of the constnt C.

157 8. Seprtion of Vribles 569 Suppose tht = () is solution to (). It then follows from the chin rule tht d d [H()]=dH d d d = h() d d = g() = dg d Since the functions H() nd G() hve the sme derivtive with respect to, the must differ b constnt (Theorem 4.8.3). It then follows tht stisfies (4) for n pproprite choice of C. Conversel, if = () is defined implicitl b Eqution (4), then implicit differentition shows tht (5) is stisfied, nd thus () is solution to () (Eercise 67). Becuse of this, it is common prctice to refer to Eqution (4) s the solution to (). In summr, we hve the following procedure for solving (), clled seprtion of vribles: (5) Seprtion of Vribles Step. Seprte the vribles in () b rewriting the eqution in the differentil form h() d = g() d Step. Integrte both sides of the eqution in Step (the left side with respect to nd the right side with respect to ): h() d = g() d Step 3. If H() is n ntiderivtive of h() nd G() is n ntiderivtive of g(), then the eqution H() = G() + C will generll define fmil of solutions implicitl. In some cses it m be possible to solve this eqution eplicitl for. For n initil-vlue problem in which the differentil eqution is seprble, ou cn either use the initil condition to solve for C, s in Emple, or replce the indefinite integrls in Step b definite integrls (Eercise 68). Emple Solve the differentil eqution d d = 4 nd then solve the initil-vlue problem d d = 4, () = Solution. For = we cn write the differentil eqution in form () s d d = 4 Seprting vribles nd integrting ields d = 4d d = 4d or = + C Solving for s function of, we obtin = C

158 57 Chpter 8 / Mthemticl Modeling with Differentil Equtions The initil condition () = requires tht = when =. Substituting these vlues into our solution ields C = (verif). Thus, solution to the initil-vlue problem is = + Some integrl curves nd our solution of the initil-vlue problem re grphed in Figure 8... (6) d Integrl curves for = 4 d Figure 8.. One spect of our solution to Emple deserves specil comment. Hd the initil condition been () = insted of () =, the method we used would hve filed to ield solution to the resulting initil-vlue problem (Eercise 5). This is due to the fct tht we ssumed = in order to rewrite the eqution d/d = 4 in the form d d = 4 It is importnt to be wre of such ssumptions when mnipulting differentil eqution lgebricll. Emple Solve the initil-vlue problem (4 cos ) d d 3 =, () = The solution of n initil-vlue problem in nd cn sometimes be epressed eplicitl s function of [s in Formul (6) of Emple ], or eplicitl s function of [s in Formul (8) of Emple ]. However, sometimes the solution cnnot be epressed in either such form, so the onl option is to epress it implicitl s n eqution in nd Integrl curves for d (4 cos ) 3 = d Figure 8.. Solution. We cn write the differentil eqution in form () s (4 cos ) d d = 3 Seprting vribles nd integrting ields or (4 cos )d = 3 d (4 cos )d = 3 d sin = 3 + C (7) For the initil-vlue problem, the initil condition () = requires tht = if =. Substituting these vlues into (7) to determine the constnt of integrtion ields C = (verif). Thus, the solution of the initil-vlue problem is or sin = 3 = 3 sin (8) Some integrl curves nd the solution of the initil-vlue problem in Emple re grphed in Figure 8... Initil-vlue problems often result from geometricl questions, s in the following emple. Emple 3 Find curve in the -plne tht psses through (, 3) nd whose tngent line t point (, ) hs slope /.

159 8. Seprtion of Vribles 57 TECHNOLOGY MASTERY Some computer lgebr sstems cn grph implicit equtions. Figure 8.. shows the grphs of (7) for C =, ±, ±, nd ±3. If ou hve CAS tht cn grph implicit equtions, tr to duplicte this figure. Solution. Since the slope of the tngent line is d/d, wehve d d = (9) nd, since the curve psses through (, 3), we hve the initil condition () = 3 Eqution (9) is seprble nd cn be written s so d = d d = d or 3 3 = + C It follows from the initil condition tht = 3if =. Substituting these vlues into the lst eqution ields C = 9 (verif), so the eqution of the desired curve is 3 3 = + 9 or = (3 + 7) /3 EXPONENTIAL GROWTH AND DECAY MODELS The popultion growth nd phrmcolog models developed in Section 8. re emples of generl clss of models clled eponentil models. In generl, eponentil models rise in situtions where quntit increses or decreses t rte tht is proportionl to the mount of the quntit present. More precisel, we mke the following definition. 8.. definition A quntit = (t) is sid to hve n eponentil growth model if it increses t rte tht is proportionl to the mount of the quntit present, nd it is sid to hve n eponentil dec model if it decreses t rte tht is proportionl to the mount of the quntit present. Thus, for n eponentil growth model, the quntit (t) stisfies n eqution of the form d = k (k > ) () dt nd for n eponentil dec model, the quntit (t) stisfies n eqution of the form d = k (k > ) () dt The constnt k is clled the growth constnt or the dec constnt, s pproprite. Equtions () nd () re seprble since the hve the form of (), but with t rther thn s the independent vrible. To illustrte how these equtions cn be solved, suppose tht positive quntit = (t) hs n eponentil growth model nd tht we know the mount of the quntit t some point in time, s = when t =. Thus, formul for (t) cn be obtined b solving the initil-vlue problem d = k, () = dt Seprting vribles nd integrting ields d = kdt or (since >) ln = kt + C ()

160 57 Chpter 8 / Mthemticl Modeling with Differentil Equtions The initil condition implies tht = when t =. Substituting these vlues in () ields C = ln (verif). Thus, from which it follows tht or, equivlentl, ln = kt + ln = e ln = e kt+ln = e kt (3) We leve it for ou to show tht if = (t) hs n eponentil dec model, nd if () =, then = e kt (4) It is stndrd prctice in pplictions to cll (5) the growth rte, even though it is misleding (the growth rte is d/dt). However, the prctice is so common tht we will follow it here. INTERPRETING THE GROWTH AND DECAY CONSTANTS The significnce of the constnt k in Formuls (3) nd (4) cn be understood b reemining the differentil equtions tht gve rise to these formuls. For emple, in the cse of the eponentil growth model, Eqution () cn be rewritten s k = d/dt (5) which sttes tht the growth rte s frction of the entire popultion remins constnt over time, nd this constnt is k. For this reson, k is clled the reltive growth rte of the popultion. It is usul to epress the reltive growth rte s percentge. Thus, reltive growth rte of 3% per unit of time in n eponentil growth model mens tht k =.3. Similrl, the constnt k in n eponentil dec model is clled the reltive dec rte. Emple 4 According to the U.S. Census Bureu, the world popultion in ws 6.9 billion nd growing t rte of bout.% per er. Assuming n eponentil growth model, estimte the world popultion t the beginning of the er 3. In Emple 4 the growth rte ws given, so there ws no need to clculte it. If the growth rte or dec rte is unknown, then it cn be clculted using the initil condition nd the vlue of t nother point in time (Eercise 44). Solution. We ssume tht the popultion t the beginning of ws 6.9 billion nd let t = time elpsed from the beginning of (in ers) = world popultion (in billions) Since the beginning of corresponds to t =, it follows from the given dt tht = () = 6.9 billion Since the growth rte is.% (k =.), it follows from (3) tht the world popultion t time t will be (t) = e kt = 6.9e.t (6) Since the beginning of the er 3 corresponds to n elpsed time of t = 9 ers (3 = 9), it follows from (6) tht the world popultion b the er 3 will be (9) = 6.9e.(9) 8.5 which is popultion of pproimtel 8.5 billion. DOUBLING TIME AND HALF-LIFE If quntit hs n eponentil growth model, then the time required for the originl size to double is clled the doubling time, nd if hs n eponentil dec model, then the time required for the originl size to reduce b hlf is clled the hlf-life. As it turns out, doubling time nd hlf-life depend onl on the growth or dec rte nd not on the mount present initill. To see wh this is so, suppose tht = (t) hs n eponentil growth model = e kt (7)

161 8. Seprtion of Vribles 573 nd let T denote the mount of time required for to double in size. Thus, t time t = T the vlue of will be, nd hence from (7) 8 = e kt or e kt = 4 Tking the nturl logrithm of both sides ields kt = ln, which implies tht the doubling time is T = ln (8) k / T T 3T Eponentil growth model with doubling time T t We leve it s n eercise to show tht Formul (8) lso gives the hlf-life of n eponentil dec model. Observe tht this formul does not involve the initil mount,so tht in n eponentil growth or dec model, the quntit doubles (or reduces b hlf) ever T units (Figure 8..3). Emple 5 It follows from (8) tht with continued growth rte of.% per er, the doubling time for the world popultion will be T = ln 63. or pproimtel 63 ers. Thus, with continued.% nnul growth rte the popultion of 6.9 billion in will double to 3.8 billion b the er 74 nd will double gin to 7.6 billion b 37. /4 /8 T T 3T Eponentil dec model with hlf-life T Figure 8..3 t RADIOACTIVE DECAY It is fct of phsics tht rdioctive elements disintegrte spontneousl in process clled rdioctive dec. Eperimenttion hs shown tht the rte of disintegrtion is proportionl to the mount of the element present, which implies tht the mount = (t) of rdioctive element present s function of time hs n eponentil dec model. Ever rdioctive element hs specific hlf-life; for emple, the hlf-life of rdioctive crbon-4 is bout 573 ers. Thus, from (8), the dec constnt for this element is k = ln ln = T 573. nd this implies tht if there re units of crbon-4 present t time t =, then the number of units present fter t ers will be pproimtel (t) = e.t (9) Emple 6 If grms of rdioctive crbon-4 re stored in cve for ers, how mn grms will be left t tht time? Solution. From (9) with = nd t =, we obtin Thus, bout 88.6 grms will be left. () = e.() = e CARBON DATING When the nitrogen in the Erth s upper tmosphere is bombrded b cosmic rdition, the rdioctive element crbon-4 is produced. This crbon-4 combines with ogen to form crbon dioide, which is ingested b plnts, which in turn re eten b nimls. In this w ll living plnts nd nimls bsorb quntities of rdioctive crbon-4. In 947 the Americn nucler scientist W. F. Libb proposed the theor tht the percentge of W. F. Libb, Rdiocrbon Dting, Americn Scientist, Vol. 44, 956, pp. 98.

162 574 Chpter 8 / Mthemticl Modeling with Differentil Equtions crbon-4 in the tmosphere nd in living tissues of plnts is the sme. When plnt or niml dies, the crbon-4 in the tissue begins to dec. Thus, the ge of n rtifct tht contins plnt or niml mteril cn be estimted b determining wht percentge of its originl crbon-4 content remins. Vrious procedures, clled crbon dting or crbon-4 dting, hve been developed for mesuring this percentge. Emple 7 In 988 the Vticn uthorized the British Museum to dte cloth relic known s the Shroud of Turin, possibl the buril shroud of Jesus of Nzreth. This cloth, which first surfced in 356, contins the negtive imge of humn bod tht ws widel believed to be tht of Jesus. The report of the British Museum showed tht the fibers in the cloth contined between 9% nd 93% of their originl crbon-4. Use this informtion to estimte the ge of the shroud. Ptrick Mesner/Liison Agenc, Inc./Gett Imges The Shroud of Turin Solution. From (9), the frction of the originl crbon-4 tht remins fter t ers is (t) = e.t Tking the nturl logrithm of both sides nd solving for t, we obtin ( ) (t) t =. ln Thus, tking (t)/ to be.93 nd.9, we obtin t = ln(.93) 6. t = ln(.9) 689. This mens tht when the test ws done in 988, the shroud ws between 6 nd 689 ers old, thereb plcing its origin between 99 A.D. nd 388 A.D. Thus, if one ccepts the vlidit of crbon-4 dting, the Shroud of Turin cnnot be the buril shroud of Jesus of Nzreth. QUICK CHECK EXERCISES 8. (See pge 579 for nswers.). Solve the first-order seprble eqution h() d d = g() b completing the following steps: Step. Seprte the vribles b writing the eqution in the differentil form. Step. Integrte both sides of the eqution in Step :. Step 3. If H() is n ntiderivtive of h(), G() is n ntiderivtive of g(), nd C is n unspecified constnt, then, s suggested b Step, the eqution will generll define fmil of solutions to h() d/d = g() implicitl.. Suppose tht quntit = (t) hs n eponentil growth model with growth constnt k>. () (t) stisfies first-order differentil eqution of the form d/dt =. (b) In terms of k, the doubling time of the quntit is. (c) If = () is the initil mount of the quntit, then n eplicit formul for (t) is given b (t) =. 3. Suppose tht quntit = (t) hs n eponentil dec model with dec constnt k>. () (t) stisfies first-order differentil eqution of the form d/dt =. (b) In terms of k, the hlf-life of the quntit is. (c) If = () is the initil mount of the quntit, then n eplicit formul for (t) is given b (t) =.

163 8. Seprtion of Vribles The initil-vlue problem d d =, () = hs solution () =. EXERCISE SET 8. Grphing Utilit C CAS C Solve the differentil eqution b seprtion of vribles. Where resonble, epress the fmil of solutions s eplicit functions of. d. d =. + d 3. + d = 5. ( + ) = e 6. = d d = ( + ) 4. ( + 4 ) d d = 3 7. e sin cos = 8. ( + )( + ) = d 9. d d =. sin d sec = 4 Solve the initil-vlue problem b seprtion of vribles.. 3 = + cos, () = π. e = e, () = 3. d = t +, dt () = 4. cosh cosh =, () = 3 5. () Sketch some tpicl integrl curves of the differentil eqution = /. (b) Find n eqution for the integrl curve tht psses through the point (, ). 6. () Sketch some tpicl integrl curves of the differentil eqution = /. (b) Find n eqution for the integrl curve tht psses through the point (3, 4). 7 8 Solve the differentil eqution nd then use grphing utilit to generte five integrl curves for the eqution. 7. ( + 4) d d + = 8. (cos ) = cos 9 Solve the differentil eqution. If ou hve CAS with implicit plotting cpbilit, use the CAS to generte five integrl curves for the eqution. 9. =. = + 4 True Flse Determine whether the sttement is true or flse. Eplin our nswer.. Ever differentil eqution of the form = f() is seprble.. A differentil eqution of the form h() d d = g() is not seprble. 3. If rdioctive element hs hlf-life of minute, nd if continer holds 3 g of the element t : p.m., then the mount remining t :5 p.m. will be g. 4. If popultion is growing eponentill, then the time it tkes the popultion to qudruple is independent of the size of the popultion. 5. Suppose tht the initil condition in Emple hd been () =. Show tht none of the solutions generted in Emple stisf this initil condition, nd then solve the initil-vlue problem d d = 4, () = Wh does the method of Emple fil to produce this prticulr solution? 6. Find ll ordered pirs (, ) such tht if the initil condition in Emple is replced b ( ) =, the solution of the resulting initil-vlue problem is defined for ll rel numbers. 7. Find n eqution of curve with -intercept whose tngent line t n point (, ) hs slope e. 8. Use grphing utilit to generte curve tht psses through the point (, ) nd whose tngent line t (, ) is perpendiculr to the line through (, ) with slope /(3 ). 9. Suppose tht n initil popultion of, bcteri grows eponentill t rte of % per hour nd tht = (t) is the number of bcteri present t hours lter. () Find n initil-vlue problem whose solution is (t). (b) Find formul for (t). (c) How long does it tke for the initil popultion of bcteri to double? (d) How long does it tke for the popultion of bcteri to rech 45,? 3. A cell of the bcterium E. coli divides into two cells ever minutes when plced in nutrient culture. Let = (t) be the number of cells tht re present t minutes fter single cell is plced in the culture. Assume tht the growth of the bcteri is pproimted b n eponentil growth model. () Find n initil-vlue problem whose solution is (t). (b) Find formul for (t). (cont.)

164 576 Chpter 8 / Mthemticl Modeling with Differentil Equtions (c) How mn cells re present fter hours? (d) How long does it tke for the number of cells to rech,,? 3. Rdon- is rdioctive gs with hlf-life of 3.83 ds. This gs is helth hzrd becuse it tends to get trpped in the bsements of houses, nd mn helth officils suggest tht homeowners sel their bsements to prevent entr of the gs. Assume tht 5. 7 rdon toms re trpped in bsement t the time it is seled nd tht (t)is the number of toms present t ds lter. () Find n initil-vlue problem whose solution is (t). (b) Find formul for (t). (c) How mn toms will be present fter 3 ds? (d) How long will it tke for 9% of the originl quntit of gs to dec? 3. Polonium- is rdioctive element with hlf-life of 4 ds. Assume tht milligrms of the element re plced in led continer nd tht (t) is the number of milligrms present t ds lter. () Find n initil-vlue problem whose solution is (t). (b) Find formul for (t). (c) How mn milligrms will be present fter weeks? (d) How long will it tke for 7% of the originl smple to dec? 33. Suppose tht fruit flies re plced in breeding continer tht cn support t most, flies. Assuming tht the popultion grows eponentill t rte of % per d, how long will it tke for the continer to rech cpcit? 34. Suppose tht the town of Grrock hd popultion of, in 6 nd popultion of, in. Assuming n eponentil growth model, in wht er will the popultion rech,? 35. A scientist wnts to determine the hlf-life of certin rdioctive substnce. She determines tht in ectl 5 ds.-milligrm smple of the substnce decs to 3.5 milligrms. Bsed on these dt, wht is the hlf-life? 36. Suppose tht 3% of certin rdioctive substnce decs in 5 ers. () Wht is the hlf-life of the substnce in ers? (b) Suppose tht certin quntit of this substnce is stored in cve. Wht percentge of it will remin fter t ers? FOCUS ON CONCEPTS 37. () Mke conjecture bout the effect on the grphs of = e kt nd = e kt of vring k nd keeping fied. Confirm our conjecture with grphing utilit. (b) Mke conjecture bout the effect on the grphs of = e kt nd = e kt of vring nd keeping k fied. Confirm our conjecture with grphing utilit. 38. () Wht effect does incresing nd keeping k fied hve on the doubling time or hlf-life of n eponentil model? Justif our nswer. (b) Wht effect does incresing k nd keeping fied hve on the doubling time nd hlf-life of n eponentil model? Justif our nswer. 39. () There is trick, clled the Rule of 7, tht cn be used to get quick estimte of the doubling time or hlf-life of n eponentil model. According to this rule, the doubling time or hlf-life is roughl 7 divided b the percentge growth or dec rte. For emple, we showed in Emple 5 tht with continued growth rte of.% per er the world popultion would double ever 63 ers. This result grees with the Rule of 7, since 7/ Eplin wh this rule works. (b) Use the Rule of 7 to estimte the doubling time of popultion tht grows eponentill t rte of % per er. (c) Use the Rule of 7 to estimte the hlf-life of popultion tht decreses eponentill t rte of 3.5% per hour. (d) Use the Rule of 7 to estimte the growth rte tht would be required for popultion growing eponentill to double ever ers. 4. Find formul for the tripling time of n eponentil growth model. 4. In 95, reserch tem digging ner Folsom, New Meico, found chrred bison bones long with some lef-shped projectile points (clled the Folsom points ) tht hd been mde b Pleo-Indin hunting culture. It ws cler from the evidence tht the bison hd been cooked nd eten b the mkers of the points, so tht crbon-4 dting of the bones mde it possible for the reserchers to determine when the hunters romed North Americ. Tests showed tht the bones contined between 7% nd 3% of their originl crbon- 4. Use this informtion to show tht the hunters lived roughl between 9 b.c. nd 8 b.c. 4. () Use grphing utilit to mke grph of p rem versus t, where p rem is the percentge of crbon-4 tht remins in n rtifct fter t ers. (b) Use the grph to estimte the percentge of crbon-4 tht would hve to hve been present in the 988 test of the Shroud of Turin for it to hve been the buril shroud of Jesus of Nzreth (see Emple 7). 43. () It is currentl ccepted tht the hlf-life of crbon-4 might vr ±4 ers from its nominl vlue of 573 ers. Does this vrition mke it possible tht the Shroud of Turin dtes to the time of Jesus of Nzreth (see Emple 7)? (b) Review the subsection of Section 3.5 entitled Error Propgtion, nd then estimte the percentge error tht

165 8. Seprtion of Vribles 577 results in the computed ge of n rtifct from n r% error in the hlf-life of crbon Suppose tht quntit hs n eponentil growth model = e kt or n eponentil dec model = e kt, nd it is known tht = if t = t. In ech cse find formul for k in terms of,, nd t, ssuming tht t =. 45. () Show tht if quntit = (t) hs n eponentil model, nd if (t ) = nd (t ) =, then the doubling time or the hlf-life T is T = (t t ) ln ln( / ) (b) In certin -hour period the number of bcteri in colon increses b 5%. Assuming n eponentil growth model, wht is the doubling time for the colon? 46. Suppose tht P dollrs is invested t n nnul interest rte of r %. If the ccumulted interest is credited to the ccount t the end of the er, then the interest is sid to be compounded nnull; if it is credited t the end of ech 6-month period, then it is sid to be compounded seminnull; nd if it is credited t the end of ech 3-month period, then it is sid to be compounded qurterl. The more frequentl the interest is compounded, the better it is for the investor since more of the interest is itself erning interest. () Show tht if interest is compounded n times er t equll spced intervls, then the vlue A of the investment fter t ers is ( A = P + r ) nt n (b) One cn imgine interest to be compounded ech d, ech hour, ech minute, nd so forth. Crried to the limit one cn conceive of interest compounded t ech instnt of time; this is clled continuous compounding. Thus, from prt (), the vlue A of P dollrs fter t ers when invested t n nnul rte of r %, compounded continuousl, is A = lim n + P ( + r n Use the fct tht lim ( + ) / = e to prove tht A = Pe rt. (c) Use the result in prt (b) to show tht mone invested t continuous compound interest increses t rte proportionl to the mount present. 47. () If $ is invested t 8% per er compounded continuousl (Eercise 46), wht will the investment be worth fter 5 ers? (b) If it is desired tht n investment t 8% per er compounded continuousl should hve vlue of $, fter ers, how much should be invested now? (c) How long does it tke for n investment t 8% per er compounded continuousl to double in vlue? ) nt 48. Wht is the effective nnul interest rte for n interest rte of r% per er compounded continuousl? 49. Assume tht = (t) stisfies the logistic eqution with = () the initil vlue of. () Use seprtion of vribles to derive the solution L = + (L )e kt (b) Use prt () to show tht lim (t) = L. t + 5. Use our nswer to Eercise 49 to derive solution to the model for the spred of disese [Eqution (6) of Section 8.]. 5. The grph of solution to the logistic eqution is known s logistic curve, nd if >, it hs one of four generl shpes, depending on the reltionship between nd L. In ech prt, ssume tht k = nd use grphing utilit to plot logistic curve stisfing the given condition. () >L (b) = L (c) L/ <L (d) < <L/ 5 53 The grph of logistic model L = + (L )e kt is shown. Estimte, L, nd k t Plot solution to the initil-vlue problem d ( =.98 ), = dt Suppose tht the growth of popultion = (t) is given b the logistic eqution 6 = 5 + 7e t () Wht is the popultion t time t =? (b) Wht is the crring cpcit L? (c) Wht is the constnt k? (d) When does the popultion rech hlf of the crring cpcit? (e) Find n initil-vlue problem whose solution is (t). 56. Suppose tht the growth of popultion = (t) is given b the logistic eqution = + 999e.9t () Wht is the popultion t time t =? (b) Wht is the crring cpcit L? (c) Wht is the constnt k? t (cont.)

166 578 Chpter 8 / Mthemticl Modeling with Differentil Equtions (d) When does the popultion rech 75% of the crring cpcit? (e) Find n initil-vlue problem whose solution is (t). 57. Suppose tht universit residence hll houses students. Following the semester brek, students in the hll return with the flu, nd 5 ds lter 35 students hve the flu. () Use the result of Eercise 5 to find the number of students who will hve the flu t ds fter returning to school. (b) Mke tble tht illustrtes how the flu spreds d to d over -week period. (c) Use grphing utilit to generte grph tht illustrtes how the flu spreds over -week period. 58. Suppose tht t time t = n object with temperture T is plced in room with constnt temperture T.IfT <T, then the temperture of the object will increse, nd if T >T, then the temperture will decrese. Assuming tht Newton s Lw of Cooling pplies, show tht in both cses the temperture T(t)t time t is given b T(t)= T + (T T )e kt where k is positive constnt. 59. A cup of wter with temperture of 95 C is plced in room with constnt temperture of C. () Assuming tht Newton s Lw of Cooling pplies, use the result of Eercise 58 to find the temperture of the wter t minutes fter it is plced in the room. [Note: The solution will involve constnt of proportionlit.] (b) How mn minutes will it tke for the wter to rech temperture of 5 C if it cools to 85 C in minute? 6. A glss of lemonde with temperture of 4 F is plced in room with constnt temperture of 7 F, nd hour lter its temperture is 5 F. Show tht t hours fter the lemonde is plced in the room its temperture is pproimted b T = 7 3e.5t. 6. A rocket, fired upwrd from rest t time t =, hs n initil mss of m (including its fuel). Assuming tht the fuel is consumed t constnt rte k, the mss m of the rocket, while fuel is being burned, will be given b m = m kt. It cn be shown tht if ir resistnce is neglected nd the fuel gses re epelled t constnt speed c reltive to the rocket, then the velocit v of the rocket will stisf the eqution m dv = ck mg dt where g is the ccelertion due to grvit. () Find v(t) keeping in mind tht the mss m is function of t. (b) Suppose tht the fuel ccounts for 8% of the initil mss of the rocket nd tht ll of the fuel is consumed in s. Find the velocit of the rocket in meters per second t the instnt the fuel is ehusted. [Note: Tke g = 9.8m/s nd c = 5 m/s.] 6. A bullet of mss m, fired stright up with n initil velocit of v, is slowed b the force of grvit nd drg force of ir resistnce kv, where k is positive constnt. As the bullet moves upwrd, its velocit v stisfies the eqution m dv = (kv + mg) dt where g is the constnt ccelertion due to grvit. () Show tht if = (t) is the height of the bullet bove the brrel opening t time t, then mv dv d = (kv + mg) (b) Epress in terms of v given tht = when v = v. (c) Assuming tht v = 988 m/s, g = 9.8 m/s m = kg, k = kg/m use the result in prt (b) to find out how high the bullet rises. [Hint: Find the velocit of the bullet t its highest point.] Suppose tht tnk contining liquid is vented to the ir t the top nd hs n outlet t the bottom through which the liquid cn drin. It follows from Torricelli s lw in phsics tht if the outlet is opened t time t =, then t ech instnt the depth of the liquid h(t) nd the re A(h) of the liquid s surfce re relted b A(h) dh dt = k h where k is positive constnt tht depends on such fctors s the viscosit of the liquid nd the cross-sectionl re of the outlet. Use this result in these eercises, ssuming tht h is in feet, A(h) is in squre feet, nd t is in seconds. 63. Suppose tht the clindricl tnk in the ccompning figure is filled to depth of 4 feet t time t = nd tht the constnt in Torricelli s lw is k =.5. () Find h(t). (b) How mn minutes will it tke for the tnk to drin completel? 64. Follow the directions of Eercise 63 for the clindricl tnk in the ccompning figure, ssuming tht the tnk is filled to depth of 4 feet t time t = nd tht the constnt in Torricelli s lw is k =.5. 4 ft ft Figure E-63 4 ft 6 ft Figure E-64

167 8.3 Slope Fields; Euler s Method Suppose tht prticle moving long the -is encounters resisting force tht results in n ccelertion of = dv/dt = 3 v. If = cm nd v = 8 cm/s t time t =, find the velocit v nd position s function of t for t. 66. Suppose tht prticle moving long the -is encounters resisting force tht results in n ccelertion of = dv/dt =. v. Given tht = cm nd v = 9 cm/s t time t =, find the velocit v nd position s function of t for t. FOCUS ON CONCEPTS 67. Use implicit differentition to prove tht n differentible function defined implicitl b Eqution (4) will be solution to (). 68. Prove tht solution to the initil-vlue problem h() d d = g(), ( ) = is defined implicitl b the eqution h(r) dr = g(s) ds 69. Let L denote tngent line t (, ) to solution of Eqution (), nd let (, ), (, ) denote n two points on L. Prove tht Eqution () is stisfied b d = = nd d = =. 7. Writing A student objects to the method of seprtion of vribles becuse it often produces n eqution in nd insted of n eplicit function = f(). Discuss the pros nd cons of this student s position. 7. Writing A student objects to Step in the method of seprtion of vribles becuse one side of the eqution is integrted with respect to while the other side is integrted with respect to. Answer this student s objection. [Hint: Recll the method of integrtion b substitution.] QUICK CHECK ANSWERS 8.. Step : h() d = g() d; Step : h() d = 3. () k (b) ln (c) e kt 4. = k g() d; Step 3: H() = G() + C. () k (b) ln k (c) e kt 8.3 SLOPE FIELDS; EULER S METHOD In this section we will reemine the concept of slope field nd we will discuss method for pproimting solutions of first-order equtions numericll. Numericl pproimtions re importnt in cses where the differentil eqution cnnot be solved ectl. FUNCTIONS OF TWO VARIABLES We will be concerned here with first-order equtions tht re epressed with the derivtive b itself on one side of the eqution. For emple, = 3 nd = sin() In pplied problems involving time, it is usul to use t s the independent vrible, in which cse one would be concerned with equtions of the form = f(t,), where = d/dt. The first of these equtions involves onl on the right side, so it hs the form = f(). However, the second eqution involves both nd on the right side, so it hs the form = f(,), where the smbol f(,) stnds for function of the two vribles nd. Lter in the tet we will stud functions of two vribles in more depth, but for now it will suffice to think of f(,) s formul tht produces unique output when vlues of nd re given s inputs. For emple, if f(,) = + 3

168 58 Chpter 8 / Mthemticl Modeling with Differentil Equtions nd if the inputs re = nd = 4, then the output is f(, 4) = + 3( 4) = 4 = 8 Slope = f (, ) (, ) At ech point (, ) on n integrl curve of = f (, ), the tngent line hs slope f (, ). Figure 8.3. SLOPE FIELDS In Section 5. we introduced the concept of slope field in the contet of differentil equtions of the form = f(); the sme principles ppl to differentil equtions of the form = f(,) To see wh this is so, let us review the bsic ide. If we interpret s the slope of tngent line, then the differentil eqution sttes tht t ech point (, ) on n integrl curve, the slope of the tngent line is equl to the vlue of f t tht point (Figure 8.3.). For emple, suppose tht f(,) =, in which cse we hve the differentil eqution = () A geometric description of the set of integrl curves cn be obtined b choosing rectngulr grid of points in the -plne, clculting the slopes of the tngent lines to the integrl curves t the gridpoints, nd drwing smll segments of the tngent lines through those points. The resulting picture is clled slope field or direction field for the differentil eqution becuse it shows the slope or direction of the integrl curves t the gridpoints. The more gridpoints tht re used, the better the description of the integrl curves. For emple, Figure 8.3. shows two slope fields for () the first ws obtined b hnd clcultion using the 49 gridpoints shown in the ccompning tble, nd the second, which gives clerer picture of the integrl curves, ws obtined using 65 gridpoints nd CAS. vlues of f(, ) = = 3 = = = = = = 3 = = 3 = 4 3 = = = = Figure 8.3. It so hppens tht Eqution () cn be solved ectl using method we will introduce in Section 8.4. We leve it for ou to confirm tht the generl solution of this eqution is = + + Ce () Confirm tht the first slope field in Figure 8.3. is consistent with the ccompning tble in tht figure. Figure shows some of the integrl curves superimposed on the slope field. Note tht it ws not necessr to hve the generl solution to construct the slope field. Indeed, slope fields re importnt precisel becuse the cn be constructed in cses where the differentil eqution cnnot be solved ectl.

169 8.3 Slope Fields; Euler s Method 58 3 EULER S METHOD Consider n initil-vlue problem of the form Figure = f(,), ( ) = The slope field for the differentil eqution = f(,) gives us w to visulize the solution of the initil-vlue problem, since the grph of the solution is the integrl curve tht psses through the point (, ). The slope field will lso help us to develop method for pproimting the solution to the initil-vlue problem numericll. We will not ttempt to pproimte () for ll vlues of ; rther, we will choose some smll increment nd focus on pproimting the vlues of () t succession of -vlues spced units prt, strting from. We will denote these -vlues b = +, = +, 3 = +, 4 = 3 +,... nd we will denote the pproimtions of () t these points b ( ), ( ), 3 ( 3 ), 4 ( 4 ),... ( 4, 4 ) ( 3, 3 ) (, ) (, ) (, ) Δ Figure ( n+, n+ ) Slope = f ( n, n ) n+ n The technique tht we will describe for obtining these pproimtions is clled Euler s Method. Although there re better pproimtion methods vilble, mn of them use Euler s Method s strting point, so the underling concepts re importnt to understnd. The bsic ide behind Euler s Method is to strt t the known initil point (, ) nd drw line segment in the direction determined b the slope field until we rech the point (, ) with -coordinte = + (Figure 8.3.4). If is smll, then it is resonble to epect tht this line segment will not devite much from the integrl curve = (), nd thus should closel pproimte ( ). To obtin the subsequent pproimtions, we repet the process using the slope field s guide t ech step. Strting t the endpoint (, ), we drw line segment determined b the slope field until we rech the point (, ) with -coordinte = +, nd from tht point we drw line segment determined b the slope field to the point ( 3, 3 ) with -coordinte 3 = +, nd so forth. As indicted in Figure 8.3.4, this procedure produces polgonl pth tht tends to follow the integrl curve closel, so it is resonble to epect tht the -vlues, 3, 4,... will closel pproimte ( ), ( 3 ), ( 4 ),... To eplin how the pproimtions,, 3,... cn be computed, let us focus on tpicl line segment. As indicted in Figure 8.3.5, ssume tht we hve found the point ( n, n ), nd we re tring to determine the net point ( n+, n+ ), where n+ = n +. Since the slope of the line segment joining the points is determined b the slope field t the strting point, the slope is f( n, n ), nd hence ( n, n ) Figure Δ which we cn rewrite s n+ n = n+ n = f( n, n ) n+ n n+ = n + f( n, n ) This formul, which is the hert of Euler s Method, tells us how to use ech pproimtion to compute the net pproimtion.

170 58 Chpter 8 / Mthemticl Modeling with Differentil Equtions Euler s Method To pproimte the solution of the initil-vlue problem = f(,), ( ) = proceed s follows: Step. Choose nonzero number to serve s n increment or step size long the -is, nd let Step. Compute successivel = +, = +, 3 = +,... 3 = + f(, ) = + f(, ) = + f(, ). n+ = n + f( n, n ) The numbers,, 3,...in these equtions re the pproimtions of ( ), ( ), ( 3 ),... Emple Use Euler s Method with step size of. to mke tble of pproimte vlues of the solution of the initil-vlue problem over the intervl. =, () = (3) Solution. In this problem we hve f(,) =, =, nd =. Moreover, since the step size is., the -vlues t which the pproimte vlues will be obtined re =., =., 3 =.3,..., 9 =.9, = The first three pproimtions re = + f(, ) = + ( )(.) =. = + f(, ) =. + (..)(.) =.4 3 = + f(, ) =.4 + (.4.)(.) =.63 Here is w of orgnizing ll pproimtions rounded to five deciml plces: euler's method for =, () = with Δ =. n n n f( n, n )Δ n+ = n + f ( n, n )Δ

171 8.3 Slope Fields; Euler s Method 583 Observe tht ech entr in the lst column becomes the net entr in the third column. This is reminiscent of Newton s Method in which ech successive pproimtion is used to find the net. As rule of thumb, the bsolute error in n pproimtion produced b Euler s Method is proportionl to the step size. Thus, reducing the step size b hlf reduces the bsolute nd percentge errors b roughl hlf. However, reducing the step size increses the mount of computtion, thereb incresing the potentil for more roundoff error. Such mtters re discussed in courses on differentil equtions or numericl nlsis. ACCURACY OF EULER S METHOD It follows from (3) nd the initil condition () = tht the ect solution of the initilvlue problem in Emple is = + + e Thus, in this cse we cn compre the pproimte vlues of () produced b Euler s Method with deciml pproimtions of the ect vlues (Tble 8.3.). In Tble 8.3. the bsolute error is clculted s nd the percentge error s ect vlue pproimtion ect vlue pproimtion ect vlue % Tble 8.3. ect solution euler pproimtion bsolute error percentge error Notice tht the bsolute error tends to increse s moves w from QUICK CHECK EXERCISES 8.3 (See pge 586 for nswers.). Mtch ech differentil eqution with its slope field. () = (b) = e (c) = (d) = III Figure E- IV I II. The slope field for = / t the 6 gridpoints (, ), where =,,, nd =,,, is shown in

172 584 Chpter 8 / Mthemticl Modeling with Differentil Equtions the ccompning figure. Use this slope field nd geometric resoning to find the integrl curve tht psses through the point (, ). 3. When using Euler s Method on the initil-vlue problem = f(,), ( ) =, we obtin n+ from n, n, nd b mens of the formul n+ =. 4. Consider the initil-vlue problem =, () =. () Use Euler s Method with two steps to pproimte (). (b) Wht is the ect vlue of ()? Figure E- EXERCISE SET 8.3 Grphing Utilit. Sketch the slope field for = /4 t the 5 gridpoints (, ), where =,,..., nd =,,...,.. Sketch the slope field for + = t the 5 gridpoints (, ), where =,,...,4 nd =,,...,4. (e) = + (f ) = (sin )(sin ) 3. A slope field for the differentil eqution = is shown in the ccompning figure. In ech prt, sketch the grph of the solution tht stisfies the initil condition. () () = (b) () = (c) () = I II 3 Figure E-3 4. Solve the initil-vlue problems in Eercise 3, nd use grphing utilit to confirm tht the integrl curves for these solutions re consistent with the sketches ou obtined from the slope field. III IV FOCUS ON CONCEPTS 5. Use the slope field in Eercise 3 to mke conjecture bout the behvior of the solutions of = s +, nd confirm our conjecture b emining the generl solution of the eqution. 6. In prts () (f ), mtch the differentil eqution with the slope field, nd eplin our resoning. () = / (b) = / (c) = e (d) = V Figure E-6 VI

173 8.3 Slope Fields; Euler s Method Use Euler s Method with the given step size or t to pproimte the solution of the initil-vlue problem over the stted intervl. Present our nswer s tble nd s grph. 7. d/d = 3, () =, 4, =.5 8. d/d =, () =,, =.5 9. d/dt = cos, () =, t, t =.5. d/dt = e, () =, t, t =.. Consider the initil-vlue problem = sin πt, () = Use Euler s Method with five steps to pproimte (). 5 True Flse Determine whether the sttement is true or flse. Eplin our nswer.. If the grph of = f()is n integrl curve for slope field, then so is n verticl trnsltion of this grph. 3. Ever integrl curve for the slope field d/d = e is the grph of n incresing function of. 4. Ever integrl curve for the slope field d/d = e is concve up. 5. If p() is cubic polnomil in, then the slope field d/d = p() hs n integrl curve tht is horizontl line. FOCUS ON CONCEPTS 6. () Show tht the solution of the initil-vlue problem = e,() = is () = e t dt (b) Use Euler s Method with =.5 to pproimte the vlue of () = e t dt nd compre the nswer to tht produced b clculting utilit with numericl integrtion cpbilit. 7. The ccompning figure shows slope field for the differentil eqution = /. () Use the slope field to estimte ( ) for the solution tht stisfies the given initil condition () =. (b) Compre our estimte to the ect vlue of ( ). Figure E-7 8. Refer to slope field II in Quick Check Eercise. () Does the slope field pper to hve horizontl line s n integrl curve? (b) Use the differentil eqution for the slope field to verif our nswer to prt (). 9. Refer to the slope field in Eercise 3 nd consider the integrl curve through (, ). () Use the slope field to estimte where the integrl curve intersects the -is. (b) Compre our estimte in prt () with the ect vlue of the -intercept for the integrl curve.. Consider the initil-vlue problem d d =, () = () Use Euler s Method with step sizes of =.,., nd.5 to obtin three pproimtions of (). (b) Find () ectl.. A slope field of the form = f() is sid to be utonomous. () Eplin wh the tngent segments long n horizontl line will be prllel for n utonomous slope field. (b) The word utonomous mens independent. In wht sense is n utonomous slope field independent? (c) Suppose tht G() is n ntiderivtive of /[f()] nd tht C is constnt. Eplin wh n differentible function defined implicitl b G() = C will be solution to the eqution = f().. () Solve the eqution = nd show tht ever nonconstnt solution hs grph tht is everwhere concve up. (b) Eplin how the conclusion in prt () m be obtined directl from the eqution = without solving. 3. () Find slope field whose integrl curve through (, ) stisfies 3 = b differentiting this eqution implicitl. (b) Prove tht if () is n integrl curve of the slope field in prt (), then [()] 3 () will be constnt function. (c) Find n eqution tht implicitl defines the integrl curve through (, ) of the slope field in prt (). 4. () Find slope field whose integrl curve through (, ) stisfies e + e = b differentiting this eqution implicitl. (b) Prove tht if () is n integrl curve of the slope field in prt (), then e () + ()e will be constnt function. (c) Find n eqution tht implicitl defines the integrl curve through (, ) of the slope field in prt ().

174 586 Chpter 8 / Mthemticl Modeling with Differentil Equtions 5. Consider the initil-vlue problem =, () =, nd let n denote the pproimtion of () using Euler s Method with n steps. () Wht would ou conjecture is the ect vlue of lim n + n? Eplin our resoning. (b) Find n eplicit formul for n nd use it to verif our conjecture in prt (). 6. Writing Eplin the connection between Euler s Method nd the locl liner pproimtion discussed in Section Writing Given slope field, wht fetures of n integrl curve might be discussed from the slope field? Appl our ides to the slope field in Eercise 3. QUICK CHECK ANSWERS 8.3. () IV (b) III (c) I (d) II. =, > 3. n + f( n, n ) 4. ().5 (b) e 8.4 FIRST-ORDER DIFFERENTIAL EQUATIONS AND APPLICATIONS In this section we will discuss generl method tht cn be used to solve lrge clss of first-order differentil equtions. We will use this method to solve differentil equtions relted to the problems of miing liquids nd free fll retrded b ir resistnce. FIRST-ORDER LINEAR EQUATIONS The simplest first-order equtions re those tht cn be written in the form d = q() () d Such equtions cn often be solved b integrtion. For emple, if d d = 3 () then = 3 d = C is the generl solution of () on the intervl (, + ). More generll, first-order differentil eqution is clled liner if it is epressible in the form d + p() = q() (3) d Eqution () is the specil cse of (3) tht results when the function p() is identicll. Some other emples of first-order liner differentil equtions re d d + = e, d d + (sin ) + 3 =, d d + 5 = p() =, q() = e p() = sin,q() = 3 p() = 5, q() = We will ssume tht the functions p() nd q() in (3) re continuous on common intervl, nd we will look for generl solution tht is vlid on tht intervl. One method for doing this is bsed on the observtion tht if we define μ = μ() b μ = e p() d (4)

175 then Thus, 8.4 First-Order Differentil Equtions nd Applictions 587 dμ d = e p() d d d If (3) is multiplied through b μ, it becomes Combining this with (5) we hve p() d = μp() d d (μ) = μ d d + dμ d = μ d + μp() (5) d μ d + μp() = μq() d d (μ) = μq() (6) d This eqution cn be solved for b integrting both sides with respect to nd then dividing through b μ to obtin = μq() d (7) μ which is generl solution of (3) on the intervl. The function μ in (4) is clled n integrting fctor for (3), nd this method for finding generl solution of (3) is clled the method of integrting fctors. Although one could simpl memorize Formul (7), we recommend solving first-order liner equtions b ctull crring out the steps used to derive this formul: The Method of Integrting Fctors Step. Clculte the integrting fctor μ = e p() d Since n μ will suffice, we cn tke the constnt of integrtion to be zero in this step. Step. Multipl both sides of (3) b μ nd epress the result s d (μ) = μq() d Step 3. Integrte both sides of the eqution obtined in Step nd then solve for. Be sure to include constnt of integrtion in this step. Emple Solve the differentil eqution d d = e Solution. Compring the given eqution to (3), we see tht we hve first-order liner eqution with p() = nd q() = e. These coefficients re continuous on the intervl (, + ), so the method of integrting fctors will produce generl solution on this intervl. The first step is to compute the integrting fctor. This ields μ = e p() d = e ( )d = e

176 588 Chpter 8 / Mthemticl Modeling with Differentil Equtions Net we multipl both sides of the given eqution b μ to obtin which we cn rewrite s d e d e = e e d d [e ]=e Integrting both sides of this eqution with respect to we obtin Confirm tht the solution obtined in Emple grees with tht obtined b substituting the integrting fctor into Formul (7). e = e + C Finll, solving for ields the generl solution = e + Ce A differentil eqution of the form P() d + Q() = R() d cn be solved b dividing through b P()to put the eqution in the form of (3) nd then ppling the method of integrting fctors. However, the resulting solution will onl be vlid on intervls where p() = Q()/P()nd q() = R()/P()re both continuous. Emple Solve the initil-vlue problem d =, () = d Solution. This differentil eqution cn be written in the form of (3) b dividing through b. This ields d d = (8) where q() = is continuous on (, + ) nd p() = / is continuous on (, ) nd (, + ). Since we need p() nd q() to be continuous on common intervl, nd since our initil condition requires solution for =, we will find generl solution of (8) on the intervl (, + ). On this intervl we hve =, so tht p() d = Tking d = ln = ln the constnt of integrtion to be Thus, n integrting fctor tht will produce generl solution on the intervl (, + ) is μ = e p() d = e ln = e ln(/) = Multipling both sides of Eqution (8) b this integrting fctor ields or d d = [ ] d d =

177 8.4 First-Order Differentil Equtions nd Applictions 589 It is not ccidentl tht the initil-vlue problem in Emple hs unique solution. If the coefficients of (3) re continuous on n open intervl tht contins the point, then for n there will be unique solution of (3) on tht intervl tht stisfies the initil condition ( ) = [Eercise 9(b)]. Therefore, on the intervl (, + ), from which it follows tht = d = ln + C = ln + C (9) The initil condition () = requires tht = if =. Substituting these vlues into (9) nd solving for C ields C = (verif), so the solution of the initil-vlue problem is = ln + We conclude this section with some pplictions of first-order differentil equtions. MIXING PROBLEMS In tpicl miing problem, tnk is filled to specified level with solution tht contins known mount of some soluble substnce (s slt). The thoroughl stirred solution is llowed to drin from the tnk t known rte, nd t the sme time solution with known concentrtion of the soluble substnce is dded to the tnk t known rte tht m or m not differ from the drining rte. As time progresses, the mount of the soluble substnce in the tnk will generll chnge, nd the usul miing problem seeks to determine the mount of the substnce in the tnk t specified time. This tpe of problem serves s model for mn kinds of problems: dischrge nd filtrtion of pollutnts in river, injection nd bsorption of mediction in the bloodstrem, nd migrtions of species into nd out of n ecologicl sstem, for emple. 5 gl/min Figure 8.4. gl 5 gl/min Emple 3 At time t =, tnk contins 4 lb of slt dissolved in gl of wter. Suppose tht brine contining lb of slt per gllon of brine is llowed to enter the tnk t rte of 5 gl/min nd tht the mied solution is drined from the tnk t the sme rte (Figure 8.4.). Find the mount of slt in the tnk fter minutes. Solution. Let (t) be the mount of slt (in pounds) fter t minutes. We re given tht () = 4, nd we wnt to find (). We will begin b finding differentil eqution tht is stisfied b (t). To do this, observe tht d/dt, which is the rte t which the mount of slt in the tnk chnges with time, cn be epressed s d = rte in rte out () dt where rte in is the rte t which slt enters the tnk nd rte out is the rte t which slt leves the tnk. But the rte t which slt enters the tnk is rte in = ( lb/gl) (5 gl/min) = lb/min Since brine enters nd drins from the tnk t the sme rte, the volume of brine in the tnk sts constnt t gl. Thus, fter t minutes hve elpsed, the tnk contins (t) lb of slt per gl of brine, nd hence the rte t which slt leves the tnk t tht instnt is rte out = Therefore, () cn be written s d dt ( (t) lb /gl = ) (5 gl/min) = (t) lb /min or d dt + =

178 59 Chpter 8 / Mthemticl Modeling with Differentil Equtions which is first-order liner differentil eqution stisfied b (t). Since we re given tht () = 4, the function (t) cn be obtined b solving the initil-vlue problem d dt + =, () = 4 The integrting fctor for the differentil eqution is μ = e (/)dt = e t/ If we multipl the differentil eqution through b μ, then we obtin d dt (et/ ) = e t/ e t/ = e t/ dt = e t/ + C (t) = + Ce t/ () The initil condition sttes tht = 4 when t =. Substituting these vlues into () nd solving for C ields C = 96 (verif), so (t) = 96e t/ () Amount of slt (lb) = 96e t/ Time t (min) Figure 8.4. The grph shown in Figure 8.4. suggests tht (t) s t +. This mens tht over n etended period of time the mount of slt in the tnk tends towrd lb. Give n informl phsicl rgument to eplin wh this result is to be epected. The grph of () is shown in Figure At time t = the mount of slt in the tnk is Notice tht it follows from () tht () = 96e.5 8. lb lim (t) = t + for ll vlues of C, so regrdless of the mount of slt tht is present in the tnk initill, the mount of slt in the tnk will eventull stbilize t lb. This cn lso be seen geometricll from the slope field for the differentil eqution shown in Figure This slope field suggests the following: If the mount of slt present in the tnk is greter thn lb initill, then the mount of slt will decrese stedil over time towrd limiting vlue of lb; nd if the mount of slt is less thn lb initill, then it will increse stedil towrd limiting vlue of lb. The slope field lso suggests tht if the mount present initill is ectl lb, then the mount of slt in the tnk will st constnt t lb. This cn lso be seen from (), since C = in this cse (verif) t 5 5 Figure A MODEL OF FREE-FALL MOTION RETARDED BY AIR RESISTANCE In Section 5.7 we considered the free-fll model of n object moving long verticl is ner the surfce of the Erth. It ws ssumed in tht model tht there is no ir resistnce nd tht the onl force cting on the object is the Erth s grvit. Our gol here is to find model tht tkes ir resistnce into ccount. For this purpose we mke the following ssumptions: The object moves long verticl s-is whose origin is t the surfce of the Erth nd whose positive direction is up (Figure 5.7.7). At time t = the height of the object is s nd the velocit is v. The onl forces on the object re the force F G = mg of the Erth s grvit cting down nd the force F R of ir resistnce cting opposite to the direction of motion. The force F R is clled the drg force.

179 8.4 First-Order Differentil Equtions nd Applictions 59 In the cse of free-fll motion retrded b ir resistnce, the net force cting on the object is F G + F R = mg + F R nd the ccelertion is d s/dt, so Newton s Second Lw of Motion [Eqution (5) of Section 6.6] implies tht mg + F R = m d s (3) dt Eperimenttion hs shown tht the force F R of ir resistnce depends on the shpe of the object nd its speed the greter the speed, the greter the drg force. There re mn possible models for ir resistnce, but one of the most bsic ssumes tht the drg force F R is proportionl to the velocit of the object, tht is, F R = cv where c is positive constnt tht depends on the object s shpe nd properties of the ir. (The minus sign ensures tht the drg force is opposite to the direction of motion.) Substituting this in (3) nd writing d s/dt s dv/dt, we obtin mg cv = m dv dt Dividing b m nd rerrnging we obtin dv dt + c m v = g which is first-order liner differentil eqution in the unknown function v = v(t) with p(t) = c/m nd q(t) = g [see (3)]. For specific object, the coefficient c cn be determined eperimentll, so we will ssume tht m, g, nd c re known constnts. Thus, the velocit function v = v(t) cn be obtined b solving the initil-vlue problem dv dt + c m v = g, v() = v (4) Once the velocit function is found, the position function s = s(t) cn be obtined b solving the initil-vlue problem ds dt = v(t), s() = s (5) In Eercise 5 we will sk ou to solve (4) nd show tht Note tht v(t) = e ct/m ( v + mg c ) mg c lim v(t) = mg (7) t + c (verif). Thus, the speed v(t) does not increse indefinitel, s in free fll; rther, becuse of the ir resistnce, it pproches finite limiting speed v τ given b v τ = mg = mg (8) c c This is clled the terminl speed of the object, nd (7) is clled its terminl velocit. (6) REMARK Intuition suggests tht ner the limiting velocit, the velocit v(t) chnges ver slowl; tht is, dv/dt. Thus, it should not be surprising tht the limiting velocit cn be obtined informll from (4) b setting dv/dt = in the differentil eqution nd solving for v. This ields v = mg c which grees with (7). Other common models ssume tht FR = cv or, more generll, F R = cv p for some vlue of p.

180 59 Chpter 8 / Mthemticl Modeling with Differentil Equtions QUICK CHECK EXERCISES 8.4 (See pge 594 for nswers.). Solve the first-order liner differentil eqution d + p() = q() d b completing the following steps: Step. Clculte the integrting fctor μ =. Step. Multipl both sides of the eqution b the integrting fctor nd epress the result s d d [ ]= Step 3. Integrte both sides of the eqution obtined in Step nd solve for =.. An integrting fctor for is. d d + = q() 3. At time t =, tnk contins 3 oz of slt dissolved in 6 gl of wter. Then brine contining 5 oz of slt per gllon of brine is llowed to enter the tnk t rte of 3 gl/min nd the mied solution is drined from the tnk t the sme rte. Give n initil-vlue problem stisfied b the mount of slt (t) in the tnk t time t. Do not solve the problem. EXERCISE SET 8.4 Grphing Utilit 6 Solve the differentil eqution b the method of integrting fctors.. d d + 4 = e 3. d d + = 3. + = cos(e ) 4. d d + 4 = 5. ( + ) d d + = 7 Solve the initil-vlue problem. 7. d + =, d () = 8. d d =, () = 9. d =, d () = 3. d + =, dt () = 6. d d + + e = 4 True Flse Determine whether the sttement is true or flse. Eplin our nswer.. If nd re two solutions to first-order liner differentil eqution, then = + is lso solution.. If the first-order liner differentil eqution d + p() = q() d hs solution tht is constnt function, then q() is constnt multiple of p(). 3. In miing problem, we epect the concentrtion of the dissolved substnce within the tnk to pproch finite limit over time. 4. In our model for free-fll motion retrded b ir resistnce, the terminl velocit is proportionl to the weight of the flling object. 5. A slope field for the differentil eqution = is shown in the ccompning figure. In ech prt, sketch the grph of the solution tht stisfies the initil condition. () () = (b) () = (c) ( ) = Figure E-5 6. Solve the initil-vlue problems in Eercise 5, nd use grphing utilit to confirm tht the integrl curves for these solutions re consistent with the sketches ou obtined from the slope field. FOCUS ON CONCEPTS 7. Use the slope field in Eercise 5 to mke conjecture bout the effect of on the behvior of the solution of the initil-vlue problem =,() = s +, nd check our conjecture b emining the solution of the initil-vlue problem. 8. Consider the slope field in Eercise 5. () Use Euler s Method with =. to estimte ( ) for the solution tht stisfies the initil condition () =.

181 8.4 First-Order Differentil Equtions nd Applictions 593 (b) Would ou conjecture our nswer in prt () to be greter thn or less thn the ctul vlue of ( )? Eplin. (c) Check our conjecture in prt (b) b finding the ect vlue of ( ). 9. () Use Euler s Method with step size of =. to pproimte the solution of the initil-vlue problem = +, () = over the intervl. (b) Solve the initil-vlue problem ectl, nd clculte the error nd the percentge error in ech of the pproimtions in prt (). (c) Sketch the ect solution nd the pproimte solution together.. It ws stted t the end of Section 8.3 tht reducing the step size in Euler s Method b hlf reduces the error in ech pproimtion b bout hlf. Confirm tht the error in () is reduced b bout hlf if step size of =. is used in Eercise 9.. At time t =, tnk contins 5 oz of slt dissolved in 5 gl of wter. Then brine contining 4 oz of slt per gllon of brine is llowed to enter the tnk t rte of gl/min nd the mied solution is drined from the tnk t the sme rte. () How much slt is in the tnk t n rbitrr time t? (b) How much slt is in the tnk fter 5 min?. A tnk initill contins gl of pure wter. Then t time t = brine contining 5 lb of slt per gllon of brine is llowed to enter the tnk t rte of gl/min nd the mied solution is drined from the tnk t the sme rte. () How much slt is in the tnk t n rbitrr time t? (b) How much slt is in the tnk fter 3 min? 3. A tnk with gl cpcit initill contins 5 gl of wter tht is polluted with 5 lb of prticulte mtter. At time t =, pure wter is dded t rte of gl/min nd the mied solution is drined off t rte of gl/min. How much prticulte mtter is in the tnk when it reches the point of overflowing? 4. The wter in polluted lke initill contins lb of mercur slts per, gl of wter. The lke is circulr with dimeter 3 m nd uniform depth 3 m. Polluted wter is pumped from the lke t rte of gl/h nd is replced with fresh wter t the sme rte. Construct tble tht shows the mount of mercur in the lke (in lb) t the end of ech hour over -hour period. Discuss n ssumptions ou mde. [Note: Usem 3 = 64 gl.] 5. () Use the method of integrting fctors to derive solution (6) to the initil-vlue problem (4). [Note: Keep in mind tht c, m, nd g re constnts.] (b) Show tht (6) cn be epressed in terms of the terminl speed (8) s v(t) = e gt/vτ (v + v τ ) v τ (c) Show tht if s() = s, then the position function of the object cn be epressed s s(t) = s v τ t + v τ g (v + v τ )( e gt/vτ ) 6. Suppose full equipped skdiver weighing 4 lb hs terminl speed of ft/s with closed prchute nd 4 ft/s with n open prchute. Suppose further tht this skdiver is dropped from n irplne t n ltitude of, ft, flls for 5 s with closed prchute, nd then flls the rest of the w with n open prchute. () Assuming tht the skdiver s initil verticl velocit is zero, use Eercise 5 to find the skdiver s verticl velocit nd height t the time the prchute opens. [Note: Tke g = 3 ft/s.] (b) Use clculting utilit to find numericl solution for the totl time tht the skdiver is in the ir. 7. The ccompning figure is schemtic digrm of bsic RL series electricl circuit tht contins power source with time-dependent voltge of V(t)volts (V), resistor with constnt resistnce of R ohms ( ), nd n inductor with constnt inductnce of L henrs (H). If ou don t know nthing bout electricl circuits, don t worr; ll ou need to know is tht electricl theor sttes tht current of I(t) mperes (A) flows through the circuit where I(t) stisfies the differentil eqution L di + RI = V(t) dt () Find I(t) if R =,L = 5H,V is constnt V, nd I() = A. (b) Wht hppens to the current over long period of time? V(t) R L Figure E-7 8. Find I(t)for the electricl circuit in Eercise 7 if R = 6, L = 3H,V(t)= 3 sin t V, nd I() = 5 A. FOCUS ON CONCEPTS 9. () Prove tht n function = () defined b Eqution (7) will be solution to (3). (b) Consider the initil-vlue problem d d + p() = q(), ( ) = where the functions p() nd q() re both continuous on some open intervl. Using the generl solution for first-order liner eqution, prove tht this initil-vlue problem hs unique solution on the intervl.

182 594 Chpter 8 / Mthemticl Modeling with Differentil Equtions 3. () Prove tht solutions need not be unique for nonliner initil-vlue problems b finding two solutions to d =, () = d (b) Prove tht solutions need not eist for nonliner initil-vlue problems b showing tht there is no solution for d =, () = d 3. Writing Eplin wh the quntit μ in the Method of Integrting Fctors is clled n integrting fctor nd eplin its role in this method. 3. Writing Suppose tht given first-order differentil eqution cn be solved both b the method of integrting fctors nd b seprtion of vribles. Discuss the dvntges nd disdvntges of ech method. QUICK CHECK ANSWERS 8.4. Step : e p() d ; Step : μ, μq(); Step 3: μ μq() d. 3. d dt + = 5, () = 3 CHAPTER 8 REVIEW EXERCISES C CAS. Clssif the following first-order differentil equtions s seprble, liner, both, or neither. d d () 3 = sin (b) d d + = (c) d d = d (d) d + = sin(). Which of the given differentil equtions re seprble? d () d = f()g() d (b) d = f() g() (c) d = f() + g() d d (d) d = f()g() 3 5 Solve the differentil eqution b the method of seprtion of vribles. d 3. d = ( + ) 4. 3 tn d d sec = 5. ( + ) = e 6 8 Solve the initil-vlue problem b the method of seprtion of vribles. 6. = +, () = 7. 5 = ( + 4 ), () = 8. = 4 sec, (π/8) = 9. Sketch the integrl curve of = tht psses through the point (, ).. Sketch the integrl curve of = tht psses through the point (, ) nd the integrl curve tht psses through the point (, ).. Sketch the slope field for = /8 t the 5 gridpoints (, ), where =,,...,4 nd =,,...,4.. Solve the differentil eqution = /8, nd find fmil of integrl curves for the slope field in Eercise. 3 4 Use Euler s Method with the given step size to pproimte the solution of the initil-vlue problem over the stted intervl. Present our nswer s tble nd s grph. 3. d/d =, () =, 4, =.5 4. d/d = sin, () =,, =.5 5. Consider the initil-vlue problem = cos πt, () = Use Euler s Method with five steps to pproimte (). 6. Cloth found in n Egptin prmid contins 78.5% of its originl crbon-4. Estimte the ge of the cloth. 7. In ech prt, find n eponentil growth model = e kt tht stisfies the stted conditions. () = ; doubling time T = 5 (b) () = 5; growth rte.5% (c) () = ; () = (d) () = ; doubling time T = 5 8. Suppose tht n initil popultion of 5 bcteri grows eponentill t rte of % per hour nd tht = (t) is the number of bcteri present fter t hours. () Find n initil-vlue problem whose solution is (t). (b) Find formul for (t). (c) Wht is the doubling time for the popultion? (d) How long does it tke for the popultion of bcteri to rech 3,? 9 Solve the differentil eqution b the method of integrting fctors. d 9. d + 3 = d e. d + + e = 3 Solve the initil-vlue problem b the method of integrting fctors.

183 C. =, () = 3. + = 4, () = 3. cosh + sinh = cosh, () = 4. () Solve the initil-vlue problem = sin 3, () = b the method of integrting fctors, using CAS to perform n difficult integrtions. (b) Use the CAS to solve the initil-vlue problem directl, nd confirm tht the nswer is consistent with tht obtined in prt (). (c) Grph the solution. 5. A tnk contins gl of fresh wter. At time t = min, brine contining 5 oz of slt per gllon of brine is poured into the tnk t rte of gl/min, nd the mied solution is drined from the tnk t the sme rte. After 5 min tht Chpter 8 Mking Connections 595 process is stopped nd fresh wter is poured into the tnk t the rte of 5 gl/min, nd the mied solution is drined from the tnk t the sme rte. Find the mount of slt in the tnk t time t = 3 min. 6. Suppose tht room contining ft 3 of ir is free of crbon monoide. At time t = cigrette smoke contining 4% crbon monoide is introduced t the rte of. ft 3 /min, nd the well-circulted miture is vented from the room t the sme rte. () Find formul for the percentge of crbon monoide in the room t time t. (b) Etended eposure to ir contining.% crbon monoide is considered dngerous. How long will it tke to rech this level? Source: This is bsed on problem from Willim E. Boce nd Richrd C. DiPrim, Elementr Differentil Equtions, 7th ed., John Wile & Sons, New York,. CHAPTER 8 MAKING CONNECTIONS. Consider the first-order differentil eqution d d + p = q where p nd q re constnts. If = () is solution to this eqution, define u = u() = q p(). () Without solving the differentil eqution, show tht u grows eponentill s function of if p<, nd decs eponentill s function of if <p. (b) Use the result of prt () nd Equtions (3 4) of Section 8. to solve the initil-vlue problem d + = 4, () = d. Consider differentil eqution of the form d = f(+ b + c) d where f is function of single vrible. If = ()is solution to this eqution, define u = u() = + b() + c. () Find seprble differentil eqution tht is stisfied b the function u. (b) Use our nswer to prt () to solve d d = + 3. A first-order differentil eqution is homogeneous if it cn be written in the form d ( ) d = f for = where f is function of single vrible. If = () is solution to first-order homogeneous differentil eqution, define u = u() = ()/. () Find seprble differentil eqution tht is stisfied b the function u. (b) Use our nswer to prt () to solve d d = + 4. A first-order differentil eqution is clled Bernoulli eqution if it cn be written in the form d d + p() = q()n for n =, If = () is solution to Bernoulli eqution, define u = u() =[()] n. () Find first-order liner differentil eqution tht is stisfied b u. (b) Use our nswer to prt () to solve the initil-vlue problem d d =, () =

184 9 INFINITE SERIES istockphoto Perspective cretes the illusion tht the sequence of rilrod ties continues indefinitel but converges towrd single point infinitel fr w. In this chpter we will be concerned with infinite series, which re sums tht involve infinitel mn terms. Infinite series pl fundmentl role in both mthemtics nd science the re used, for emple, to pproimte trigonometric functions nd logrithms, to solve differentil equtions, to evlute difficult integrls, to crete new functions, nd to construct mthemticl models of phsicl lws. Since it is impossible to dd up infinitel mn numbers directl, one gol will be to define ectl wht we men b the sum of n infinite series. However, unlike finite sums, it turns out tht not ll infinite series ctull hve sum, so we will need to develop tools for determining which infinite series hve sums nd which do not. Once the bsic ides hve been developed we will begin to ppl our work; we will show how infinite series re used to evlute such quntities s ln, e, sin 3, nd π, how the re used to crete functions, nd finll, how the re used to model phsicl lws. 9. SEQUENCES In everd lnguge, the term sequence mens succession of things in definite order chronologicl order, size order, or logicl order, for emple. In mthemtics, the term sequence is commonl used to denote succession of numbers whose order is determined b rule or function. In this section, we will develop some of the bsic ides concerning sequences of numbers. DEFINITION OF A SEQUENCE Stted informll, n infinite sequence, or more simpl sequence, is n unending succession of numbers, clled terms. It is understood tht the terms hve definite order; tht is, there is first term, second term, third term 3, fourth term 4, nd so forth. Such sequence would tpicll be written s,, 3, 4,... where the dots re used to indicte tht the sequence continues indefinitel. Some specific emples re,, 3, 4,...,,, 3, 4,...,, 4, 6, 8,...,,,,,... Ech of these sequences hs definite pttern tht mkes it es to generte dditionl terms if we ssume tht those terms follow the sme pttern s the displed terms. However, 596

185 9. Sequences 597 such ptterns cn be deceiving, so it is better to hve rule or formul for generting the terms. One w of doing this is to look for function tht reltes ech term in the sequence to its term number. For emple, in the sequence, 4, 6, 8,... ech term is twice the term number; tht is, the nth term in the sequence is given b the formul n. We denote this b writing the sequence s, 4, 6, 8,...,n,... We cll the function f(n) = n the generl term of this sequence. Now, if we wnt to know specific term in the sequence, we need onl substitute its term number in the formul for the generl term. For emple, the 37th term in the sequence is 37 = 74. Emple In ech prt, find the generl term of the sequence. term number term term number term term number term Tble n Tble n... n n n Tble n n... (), 3, 3 4, 4 5,... (b), 4, 8, 6,... (c), 3, 3 4, 4,... (d), 3, 5, 7,... 5 Solution (). In Tble 9.., the four known terms hve been plced below their term numbers, from which we see tht the numertor is the sme s the term number nd the denomintor is one greter thn the term number. This suggests tht the nth term hs numertor n nd denomintor n +, s indicted in the tble. Thus, the sequence cn be epressed s, 3, 3 4, 4 5,..., n n +,... Solution (b). In Tble 9.., the denomintors of the four known terms hve been epressed s powers of nd the first four terms hve been plced below their term numbers, from which we see tht the eponent in the denomintor is the sme s the term number. This suggests tht the denomintor of the nth term is n, s indicted in the tble. Thus, the sequence cn be epressed s, 4, 8, 6,...,,... n Solution (c). This sequence is identicl to tht in prt (), ecept for the lternting signs. Thus, the nth term in the sequence cn be obtined b multipling the nth term in prt () b ( ) n+. This fctor produces the correct lternting signs, since its successive vlues, strting with n =, re,,,,...thus, the sequence cn be written s, 3, 3 4, 4 5,..., n ( )n+ n +,... Solution (d). In Tble 9..3, the four known terms hve been plced below their term numbers, from which we see tht ech term is one less thn twice its term number. This suggests tht the nth term in the sequence is n, s indicted in the tble. Thus, the sequence cn be epressed s, 3, 5, 7,...,n,... When the generl term of sequence,, 3,..., n,... ()

186 598 Chpter 9 / Infinite Series A sequence cnnot be uniquel determined from few initil terms. For emple, the sequence whose generl term is f(n) = 3 (3 5n + 6n n 3 ) hs, 3, nd 5 s its first three terms, but its fourth term is lso 5. is known, there is no need to write out the initil terms, nd it is common to write onl the generl term enclosed in brces. Thus, () might be written s { n } + n= or s { n } n= For emple, here re the four sequences in Emple epressed in brce nottion. sequence brce nottion 3 4 n n +,,,,...,, n + n + n= +,,,,..., n,... n n=,, 3, 4,..., ( ) n+ n ( ) n+ n +, n + n + n= +, 3, 5, 7,..., n,... {n } n= The letter n in () is clled the inde for the sequence. It is not essentil to use n for the inde; n letter not reserved for nother purpose cn be used. For emple, we might view the generl term of the sequence,, 3,...to be the kth term, in which cse we would denote this sequence s { k } + k=. Moreover, it is not essentil to strt the inde t ; sometimes it is more convenient to strt it t (or some other integer). For emple, consider the sequence,,,,... 3 One w to write this sequence is { n } + n= However, the generl term will be simpler if we think of the initil term in the sequence s the zeroth term, in which cse we cn write the sequence s { } + n n= We begn this section b describing sequence s n unending succession of numbers. Although this conves the generl ide, it is not stisfctor mthemticl definition becuse it relies on the term succession, which is itself n undefined term. To motivte precise definition, consider the sequence, 4, 6, 8,...,n,... If we denote the generl term b f(n) = n, then we cn write this sequence s f(), f(), f(3),...,f(n),... which is list of vlues of the function f(n) = n, n =,, 3,... whose domin is the set of positive integers. This suggests the following definition. 9.. definition A sequence is function whose domin is set of integers. Tpicll, the domin of sequence is the set of positive integers or the set of nonnegtive integers. We will regrd the epression { n } + n= to be n lterntive nottion for the function f(n) = n, n =,, 3,...,nd we will regrd { n } + n= to be n lterntive nottion for the function f(n) = n,n=,,, 3,...

187 9. Sequences 599 When the strting vlue for the inde of sequence is not relevnt to the discussion, it is common to use nottion such s { n } in which there is no reference to the strting vlue of n. We cn distinguish between different sequences b using different letters for their generl terms; thus, { n }, {b n }, nd {c n } denote three different sequences. GRAPHS OF SEQUENCES Since sequences re functions, it mkes sense to tlk bout the grph of sequence. For emple, the grph of the sequence {/n} + n= is the grph of the eqution =, n =,, 3,... n Becuse the right side of this eqution is defined onl for positive integer vlues of n, the grph consists of succession of isolted points (Figure 9..). This is different from the grph of =, which is continuous curve (Figure 9..b). n = n, n =,, 3,... =, Figure 9.. () (b) LIMIT OF A SEQUENCE Since sequences re functions, we cn inquire bout their limits. However, becuse sequence { n } is onl defined for integer vlues of n, the onl limit tht mkes sense is the limit of n s n +. In Figure 9.. we hve shown the grphs of four sequences, ech of which behves differentl s n + : The terms in the sequence {n + } increse without bound. The terms in the sequence {( ) n+ } oscillte between nd. The terms in the sequence {n/(n + )} increse towrd limiting vlue of. The terms in the sequence { + ( ) n } lso tend towrd limiting vlue of, but do so in n oscilltor fshion n n n n n + + n= n+ + ( ) n= n n + + n= + n + n= Figure 9.. Informll speking, the limit of sequence { n } is intended to describe how n behves s n +. To be more specific, we will s tht sequence { n } pproches limit L if the terms in the sequence eventull become rbitrril close to L. Geometricll, this

188 6 Chpter 9 / Infinite Series Figure 9..3 L N = L + e = L e n From this point on, the terms in the sequence re ll within e units of L. mens tht for n positive number ɛ there is point in the sequence fter which ll terms lie between the lines = L ɛ nd = L + ɛ (Figure 9..3). The following definition mkes these ides precise. How would ou define these limits? lim n + n =+ lim n + n = 9.. definition A sequence { n } is sid to converge to the limit L if given n ɛ>, there is positive integer N such tht n L <ɛfor n N. In this cse we write lim n = L n + A sequence tht does not converge to some finite limit is sid to diverge. Emple The first two sequences in Figure 9.. diverge, nd the second two converge to ; tht is, n [ ( lim = nd lim + n ] n + n + n + ) = The following theorem, which we stte without proof, shows tht the fmilir properties of limits ppl to sequences. This theorem ensures tht the lgebric techniques used to find limits of the form lim cn lso be used for limits of the form lim. + n + Additionl limit properties follow from those in Theorem For emple, use prt (e) to show tht if n L nd m is positive integer, then lim ( n) m = L m n theorem Suppose tht the sequences { n } nd {b n } converge to limits L nd L, respectivel, nd c is constnt. Then: () lim n + (b) lim n = c lim n = cl n + n + (c) lim n + b n ) = lim n + lim n = L + L n + n + n + (d ) lim n b n ) = lim n lim n = L L n + n + n + (e) lim nb n ) = lim n lim n = L L n + n + n + ) ( f ) lim n + ( n b n = lim n n + lim b = L (if L = ) n L n + If the generl term of sequence is f(n), where f() is function defined on the entire intervl [, + ), then the vlues of f(n) cn be viewed s smple vlues of f() tken

189 9. Sequences 6 L f(3) f() f() L If f () L s +, then f (n) L s n +. () (b) f () f(n) L s n +, but f() diverges b oscilltion s +. Figure 9..4 t the positive integers. Thus, if f() L s +, then f(n) L s n + (Figure 9..4). However, the converse is not true; tht is, one cnnot infer tht f() L s + from the fct tht f(n) L s n + (Figure 9..4b). Emple 3 In ech prt, determine whether the sequence converges or diverges b emining the limit s n +. { } n + { } () (b) ( ) n+ n + n + n= n + n= { (c) ( ) n+ } + (d) {8 n} + n= n Solution (). lim n + n= Dividing numertor nd denomintor b n nd using Theorem 9..3 ields n n + = lim n + = + = Thus, the sequence converges to. lim + /n = lim ( + /n) = n + n + lim n + lim + lim /n n + n + Solution (b). This sequence is the sme s tht in prt (), ecept for the fctor of ( ) n+, which oscilltes between + nd. Thus, the terms in this sequence oscillte between positive nd negtive vlues, with the odd-numbered terms being identicl to those in prt () nd the even-numbered terms being the negtives of those in prt (). Since the sequence in prt () hs limit of, it follows tht the odd-numbered terms in this sequence pproch, nd the even-numbered terms pproch. Therefore, this sequence hs no limit it diverges. Solution (c). Since /n, the product ( ) n+ (/n) oscilltes between positive nd negtive vlues, with the odd-numbered terms pproching through positive vlues nd the even-numbered terms pproching through negtive vlues. Thus, so the sequence converges to. lim n + ( )n+ n = Solution (d). lim (8 n) =, so the sequence {8 n}+ n= diverges. n + Emple 4 In ech prt, determine whether the sequence converges, nd if so, find its limit. (),,,,..., 3,... (b),, n, 3,..., n,... Solution. Replcing n b in the first sequence produces the power function (/), nd replcing n b in the second sequence produces the power function. Now recll tht if <b<, then b s +, nd if b>, then b + s + (Figure.5.).

190 6 Chpter 9 / Infinite Series Thus, lim = nd lim n + n n + n =+ So, the sequence {/ n } converges to, but the sequence { n } diverges. Emple 5 { n } + Find the limit of the sequence. e n n= Solution. The epression lim n + is n indeterminte form of tpe /, so L Hôpitl s rule is indicted. However, we cnnot ppl this rule directl to n/e n becuse the functions n nd e n hve been defined here onl t the positive integers, nd hence re not differentible functions. To circumvent this problem we etend the domins of these functions to ll rel numbers, here implied b replcing n b, nd ppl L Hôpitl s rule to the limit of the quotient /e. This ields from which we cn conclude tht lim + e = lim n + n e n lim + n e n = e = Emple 6 Solution. Show tht lim n n =. n + lim n n = n + lim n + n/n = lim n + e(/n) ln n = e = B L Hôpitl s rule pplied to (/)ln Sometimes the even-numbered nd odd-numbered terms of sequence behve sufficientl differentl tht it is desirble to investigte their convergence seprtel. The following theorem, whose proof is omitted, is helpful for tht purpose theorem A sequence converges to limit L if nd onl if the sequences of even-numbered terms nd odd-numbered terms both converge to L. L {b n } { n } {c n } Emple 7 The sequence, 3,, 3,, 3 3,... 3 converges to, since the even-numbered terms nd the odd-numbered terms both converge to, nd the sequence,,, 3,, 4,... diverges, since the odd-numbered terms converge to nd the even-numbered terms convergeto. If n L nd c n L, then b n L. Figure 9..5 THE SQUEEZING THEOREM FOR SEQUENCES The following theorem, illustrted in Figure 9..5, is n dpttion of the Squeezing Theorem (.6.4) to sequences. This theorem will be useful for finding limits of sequences tht cnnot be obtined directl. The proof is omitted.

191 9. Sequences theorem (The Squeezing Theorem for Sequences) Let { n }, {b n }, nd {c n } be sequences such tht n b n c n ( for ll vlues of n beond some inde N) If the sequences { n } nd {c n } hve common limit L s n +, then {b n } lso hs the limit L s n +. Recll tht if n is positive integer, then n! (red n fctoril ) is the product of the first n positive integers. In ddition, it is convenient to define! =. Emple 8 Use numericl evidence to mke conjecture bout the limit of the sequence { } n! + n n n= nd then confirm tht our conjecture is correct. n Tble 9..4 n! n n Solution. Tble 9..4, which ws obtined with clculting utilit, suggests tht the limit of the sequence m be. To confirm this we need to emine the limit of n = n! n n s n +. Although this is n indeterminte form of tpe /, L Hôpitl s rule is not helpful becuse we hve no definition of! for vlues of tht re not integers. However, let us write out some of the initil terms nd the generl term in the sequence: =, = =, 3 = = 9 < 3, 4 = = 3 3 < 4,... If n>, the generl term of the sequence cn be rewritten s n = 3 n n n n n = ( ) 3 n n n n n from which it follows tht n /n (wh?). It is now evident tht n n However, the two outside epressions hve limit of s n + ; thus, the Squeezing Theorem for Sequences implies tht n sn +, which confirms our conjecture. The following theorem is often useful for finding the limit of sequence with both positive nd negtive terms it sttes tht if the sequence { n } tht is obtined b tking the bsolute vlue of ech term in the sequence { n } converges to, then { n } lso converges to theorem If lim n + n =, then lim n + n =. proof Depending on the sign of n, either n = n or n = n. Thus, in ll cses we hve n n n However, the limit of the two outside terms is, nd hence the limit of n isbthe Squeezing Theorem for Sequences.

192 64 Chpter 9 / Infinite Series Emple 9 Consider the sequence,,, 3,..., ( )n n,... If we tke the bsolute vlue of ech term, we obtin the sequence,,, 3,..., n,... which, s shown in Emple 4, converges to. Thus, from Theorem 9..6 we hve [( ) n ] = n lim n + SEQUENCES DEFINED RECURSIVELY Some sequences do not rise from formul for the generl term, but rther from formul or set of formuls tht specif how to generte ech term in the sequence from terms tht precede it; such sequences re sid to be defined recursivel, nd the defining formuls re clled recursion formuls. A good emple is the mechnic s rule for pproimting squre roots. In Eercise 5 of Section 4.7 ou were sked to show tht =, n+ = ) ( n + n () describes the sequence produced b Newton s Method to pproimte s zero of the function f() =. Tble 9..5 shows the first five terms in n ppliction of the mechnic s rule to pproimte. Tble 9..5 n =, n+ = n + n deciml pproimtion = (Strting vlue). = + = = 3 + = 3/ = = 7/ 48 5 = ,857 + = /48 47,83 6 = 665,857 + = 47,83 665,857/47,83 886,73,88,897 67,3,566, It would tke us too fr field to investigte the convergence of sequences defined recursivel, but we will conclude this section with useful technique tht cn sometimes be used to compute limits of such sequences. Emple is. Assuming tht the sequence in Tble 9..5 converges, show tht the limit

193 9. Sequences 65 Solution. Assume tht n L, where L is to be determined. Since n + + s n +, it is lso true tht n+ L s n +. Thus, if we tke the limit of the epression n+ = ) ( n + n s n +, we obtin L = ( L + ) L which cn be rewritten s L =. The negtive solution of this eqution is etrneous becuse n > for ll n, sol =. QUICK CHECK EXERCISES 9. (See pge 67 for nswers.). Consider the sequence 4, 6, 8,,,... () If { n } + n= denotes this sequence, then =, 4 =, nd 7 =. The generl term is n =. (b) If {b n } + n= denotes this sequence, then b =, b 4 =, nd b 8 =. The generl term is b n =.. Wht does it men to s tht sequence { n } converges? 3. Consider sequences { n } nd {b n }, where n sn + nd b n = ( ) n. Determine which of the following sequences converge nd which diverge. If sequence converges, indicte its limit. () {b n } (b) {3 { n } } (c) {b { n } } bn (d) { n + b n } (e) n + 3 (f ) 4. Suppose tht { n }, {b n }, nd {c n } re sequences such tht n b n c n for ll n, nd tht { n } nd {c n } both converge to. Then the Theorem for Sequences implies tht {b n } converges to. EXERCISE SET 9. Grphing Utilit. In ech prt, find formul for the generl term of the sequence, strting with n =. (), 3, 9, 7,... (b), 3, 9, 7,... (c), 3 4, 5 6, 7 8,... (d) π, 4 3 π, 9 4 π, 6 5 π,.... In ech prt, find two formuls for the generl term of the sequence, one strting with n = nd the other with n =. (), r, r, r 3,... (b) r, r,r 3, r 4, () Write out the first four terms of the sequence { + ( ) n }, strting with n =. (b) Write out the first four terms of the sequence {cos nπ}, strting with n =. (c) Use the results in prts () nd (b) to epress the generl term of the sequence 4,, 4,,...in two different ws, strting with n =. 4. In ech prt, find formul for the generl term using fctorils nd strting with n =. (), 3 4, , ,... (b), 3, 3 4 5, ,... ( π ) 5 6 Let f be the function f() = cos nd define sequences { n } nd {b n } b n = f(n) nd b n = f(n + ). 5. () Does lim + f() eist? Eplin. (b) Evlute,, 3, 4, nd 5. (c) Does { n } converge? If so, find its limit. 6. () Evlute b,b,b 3,b 4, nd b 5. (b) Does {b n } converge? If so, find its limit. (c) Does {f(n)} converge? If so, find its limit. 7 Write out the first five terms of the sequence, determine whether the sequence converges, nd if so find its limit. { } n + { n } {} + n= n + n= n + n= { ( )} + { } ln n + {. ln.. n sin π } + n n= n n= n n= { ( ) 3. { + ( ) n } + n+ } + n= {( ) n n 3 n { (n + )(n + ) n } + n= } + n= n { n n } + { π n 4 n n= } + n= n= 9. {n e n } + n=. { n + 3n n} + n=

194 66 Chpter 9 / Infinite Series. {( ) n + 3 n } + n + n=. {( ) n } + n n= 3 3 Find the generl term of the sequence, strting with n =, determine whether the sequence converges, nd if so find its limit. 3., 3 4, 5 6, 7 8,... 4.,, 3, 3 4, , 9, 7,,... 6.,, 3, 4, 5,... ( 8 7. ) (, 3 ) (, 3 ) (, 4 5 ), , 3, 3, 3, ( 3 ), ( 3 4 ), ( 4 5 ), , 5 3, 6 3, 7 3, True Flse Determine whether the sttement is true or flse. Eplin our nswer. 3. Sequences re functions. 3. If { n } nd {b n } re sequences such tht { n + b n } converges, then { n } nd {b n } converge. 33. If { n } diverges, then n + or n. 34. If the grph of = f() hs horizontl smptote s +, then the sequence {f(n)} converges Use numericl evidence to mke conjecture bout the limit of the sequence, nd then use the Squeezing Theorem for Sequences (Theorem 9..5) to confirm tht our conjecture is correct. 35. lim n + sin n n FOCUS ON CONCEPTS 36. lim n + ( + n n 37. Give two emples of sequences, ll of whose terms re between nd, tht do not converge. Use grphs of our sequences to eplin their properties. 38. () Suppose tht f stisfies lim + f() =+. Is it possible tht the sequence {f(/n)} converges? Eplin. (b) Find function f such tht lim + f() does not eist but the sequence {f(/n)} converges. 39. () Strting with n =, write out the first si terms of the sequence { n }, where {, if n is odd n = n, if n is even (b) Strting with n =, nd considering the even nd odd terms seprtel, find formul for the generl term of the sequence,, 3,, 5, 4,... 6 ) n (c) Strting with n =, nd considering the even nd odd terms seprtel, find formul for the generl term of the sequence, 3, 3, 5, 5, 7, 7, 9, 9,... (d) Determine whether the sequences in prts (), (b), nd (c) converge. For those tht do, find the limit. 4. For wht positive vlues of b does the sequence b,,b,, b 3,,b 4,...converge? Justif our nswer. 4. Assuming tht the sequence given in Formul () of this section converges, use the method of Emple to show tht the limit of this sequence is. 4. Consider the sequence = 6 = = = () Find recursion formul for n+. (b) Assuming tht the sequence converges, use the method of Emple to find the limit. 43. () A bored student enters the number.5 in clcultor displ nd then repetedl computes the squre of the number in the displ. Tking =.5, find formul for the generl term of the sequence { n } of numbers tht pper in the displ. (b) Tr this with clcultor nd mke conjecture bout the limit of n. (c) Confirm our conjecture b finding the limit of n. (d) For wht vlues of will this procedure produce convergent sequence? 44. Let {, <.5 f() =,.5 < Does the sequence f(.), f(f(.)), f(f(f(.))),... converge? Justif our resoning. 45. () Use grphing utilit to generte the grph of the eqution = ( + 3 ) /, nd then use the grph to mke conjecture bout the limit of the sequence {( n + 3 n ) /n } + n= (b) Confirm our conjecture b clculting the limit. 46. Consider the sequence { n } + n= whose nth term is n = n n + (k/n) k= Show tht lim n + n = ln b interpreting n s the Riemnn sum of definite integrl.

195 9. Monotone Sequences The sequence whose terms re,,, 3, 5, 8, 3,,...is clled the Fiboncci sequence in honor of the Itlin mthemticin Leonrdo ( Fiboncci ) d Pis (c. 7 5). This sequence hs the propert tht fter strting with two s, ech term is the sum of the preceding two. () Denoting the sequence b { n } nd strting with = nd =, show tht n+ = + n if n n+ n+ (b) Give resonble informl rgument to show tht if the sequence { n+ / n } converges to some limit L, then the sequence { n+ / n+ } must lso converge to L. (c) Assuming tht the sequence { n+ / n } converges, show tht its limit is ( + 5 )/. 48. If we ccept the fct tht the sequence {/n} + n= converges to the limit L =, then ccording to Definition 9.., for ever ɛ> there eists positive integer N such tht n L = (/n) <ɛwhen n N. In ech prt, find the smllest possible vlue of N for the given vlue of ɛ. () ɛ =.5 (b) ɛ =. (c) ɛ =. 49. If we ccept the fct tht the sequence { } n + n + n= converges to the limit L =, then ccording to Definition 9.., for ever ɛ>there eists n integer N such tht n L = n n + <ɛ when n N. In ech prt, find the smllest vlue of N for the given vlue of ɛ. () ɛ =.5 (b) ɛ =. (c) ɛ =. 5. Use Definition 9.. to prove tht () the sequence {/n} + n= converges to { } n + (b) the sequence converges to. n + n= 5. Writing Discuss, with emples, vrious ws tht sequence could diverge. 5. Writing Discuss the convergence of the sequence {r n } considering the cses r <, r >, r =, nd r = seprtel. QUICK CHECK ANSWERS 9.. () 4; ; 6; n + (b) 4; ; ; n + 4. lim n eists n + 3. () diverges (b) converges to 5 (c) converges to (d) diverges (e) converges to 7 (f ) diverges 4. Squeezing; 9. MONOTONE SEQUENCES There re mn situtions in which it is importnt to know whether sequence converges, but the vlue of the limit is not relevnt to the problem t hnd. In this section we will stud severl techniques tht cn be used to determine whether sequence converges. TERMINOLOGY We begin with some terminolog. Note tht n incresing sequence need not be strictl incresing, nd decresing sequence need not be strictl decresing. 9.. definition A sequence { n } + n= is clled strictl incresing if < < 3 < < n < incresing if 3 n strictl decresing if > > 3 > > n > decresing if 3 n A sequence tht is either incresing or decresing is sid to be monotone, nd sequence tht is either strictl incresing or strictl decresing is sid to be strictl monotone. Some emples re given in Tble 9.. nd their corresponding grphs re shown in Figure 9... The first nd second sequences in Tble 9.. re strictl monotone; the third

196 68 Chpter 9 / Infinite Series nd fourth sequences re monotone but not strictl monotone; nd the fifth sequence is neither strictl monotone nor monotone. Tble 9.. sequence,, 3,..., n, n +,,,...,,... 3 n,,,, 3, 3,...,,,,,, ,,,,..., ( ) n+, n description Strictl incresing Strictl decresing Incresing; not strictl incresing Decresing; not strictl decresing Neither incresing nor decresing n n n n n + + n= n + n=,,,, 3, 3,... Cn sequence be both incresing nd decresing? Eplin n ,,,, 3, 3,... ( ) n+ n + n= n Figure 9.. TESTING FOR MONOTONICITY Frequentl, one cn guess whether sequence is monotone or strictl monotone b writing out some of the initil terms. However, to be certin tht the guess is correct, one must give precise mthemticl rgument. Tble 9.. provides two ws of doing this, one bsed Tble 9.. difference between successive terms n+ n > n+ n < n+ n n+ n rtio of successive terms n+/ n > n+/ n < n+/ n n+/ n conclusion Strictl incresing Strictl decresing Incresing Decresing

197 Figure n n + + n= n 9. Monotone Sequences 69 on differences of successive terms nd the other on rtios of successive terms. It is ssumed in the ltter cse tht the terms re positive. One must show tht the specified conditions hold for ll pirs of successive terms. Emple Use differences of successive terms to show tht, 3, 3 4,..., n n +,... (Figure 9..) is strictl incresing sequence. Solution. The pttern of the initil terms suggests tht the sequence is strictl incresing. To prove tht this is so, let n = n n + We cn obtin n+ b replcing n b n + in this formul. This ields Thus, for n n+ n = n + n + n+ = n + (n + ) + = n + n + n n + = n + n + n n = (n + )(n + ) which proves tht the sequence is strictl incresing. (n + )(n + ) > Emple Use rtios of successive terms to show tht the sequence in Emple is strictl incresing. Solution. As shown in the solution of Emple, n = n n + Forming the rtio of successive terms we obtin nd n+ = n + n + n+ = (n + ) /(n + ) = n + n n/(n + ) n + n + = n + n + n n + n from which we see tht n+ / n > for n. This proves tht the sequence is strictl incresing. () The following emple illustrtes still third technique for determining whether sequence is strictl monotone. Emple 3 In Emples nd we proved tht the sequence, 3, 3 4,..., n n +,... is strictl incresing b considering the difference nd rtio of successive terms. Alterntivel, we cn proceed s follows. Let f() = + so tht the nth term in the given sequence is n = f(n). The function f is incresing for since f ( + )() () () = = ( + ) ( + ) >

198 6 Chpter 9 / Infinite Series derivtive of f for f () > f () < f () f () Tble 9..3 conclusion for the sequence with n = f (n) Strictl incresing Strictl decresing Incresing Decresing Thus, n = f(n) < f(n + ) = n+ which proves tht the given sequence is strictl incresing. In generl, if f(n) = n is the nth term of sequence, nd if f is differentible for, then the results in Tble 9..3 cn be used to investigte the monotonicit of the sequence. PROPERTIES THAT HOLD EVENTUALLY Sometimes sequence will behve errticll t first nd then settle down into definite pttern. For emple, the sequence 9, 8, 7,,,, 3, 4,... () is strictl incresing from the fifth term on, but the sequence s whole cnnot be clssified s strictl incresing becuse of the errtic behvior of the first four terms. To describe such sequences, we introduce the following terminolog. 9.. definition If discrding finitel mn terms from the beginning of sequence produces sequence with certin propert, then the originl sequence is sid to hve tht propert eventull. 3 Figure n /n! + n= n For emple, lthough we cnnot s tht sequence () is strictl incresing, we cn s tht it is eventull strictl incresing. { n } + Emple 4 Show tht the sequence is eventull strictl decresing. n! Solution. We hve n = n n! n= nd n+ = n+ (n + )! so n+ = n+ /(n + )! = n+ n! n n /n! n (n + )! = n! (n + )n! = (3) n + From (3), n+ / n < for ll n, so the sequence is eventull strictl decresing, s confirmed b the grph in Figure AN INTUITIVE VIEW OF CONVERGENCE Informll stted, the convergence or divergence of sequence does not depend on the behvior of its initil terms, but rther on how the terms behve eventull. For emple, the sequence 3, 9, 3, 7,,, 3, 4,... eventull behves like the sequence nd hence hs limit of.,, 3,..., n,... CONVERGENCE OF MONOTONE SEQUENCES The following two theorems, whose proofs re discussed t the end of this section, show tht monotone sequence either converges or becomes infinite divergence b oscilltion cnnot occur.

199 9. Monotone Sequences theorem If sequence { n } is eventull incresing, then there re two possibilities: () There is constnt M, clled n upper bound for the sequence, such tht n M for ll n, in which cse the sequence converges to limit L stisfing L M. (b) No upper bound eists, in which cse lim n =+. n + Theorems 9..3 nd 9..4 re emples of eistence theorems; the tell us whether limit eists, but the do not provide method for finding it theorem If sequence { n } is eventull decresing, then there re two possibilities: () There is constnt M, clled lower bound for the sequence, such tht n M for ll n, in which cse the sequence converges to limit L stisfing L M. (b) No lower bound eists, in which cse lim n =. n + Emple 5 { n Show tht the sequence n! } + n= converges nd find its limit. Solution. We showed in Emple 4 tht the sequence is eventull strictl decresing. Since ll terms in the sequence re positive, it is bounded below b M =, nd hence Theorem 9..4 gurntees tht it converges to nonnegtive limit L. However, the limit is not evident directl from the formul n /n! for the nth term, so we will need some ingenuit to obtin it. It follows from Formul (3) of Emple 4 tht successive terms in the given sequence re relted b the recursion formul n+ = n + n (4) where n = n /n!. We will tke the limit s n + of both sides of (4) nd use the fct tht lim n+ = lim n = L n + n + We obtin so tht L = lim n+ = n + ( ) lim n + n + n = lim n + L = n lim = n + n! n + lim n = L = n + In the eercises we will show tht the technique illustrted in the lst emple cn be dpted to obtin n lim n + n! = (5) for n rel vlue of (Eercise 3). This result will be useful in our lter work. THE COMPLETENESS AXIOM In this tet we hve ccepted the fmilir properties of rel numbers without proof, nd indeed, we hve not even ttempted to define the term rel number. Although this is sufficient for mn purposes, it ws recognized b the lte nineteenth centur tht the stud of limits

200 6 Chpter 9 / Infinite Series nd functions in clculus requires precise iomtic formultion of the rel numbers nlogous to the iomtic development of Eucliden geometr. Although we will not ttempt to pursue this development, we will need to discuss one of the ioms bout rel numbers in order to prove Theorems 9..3 nd But first we will introduce some terminolog. If S is nonempt set of rel numbers, then we cll u n upper bound for S if u is greter thn or equl to ever number in S, nd we cll l lower bound for S if l is smller thn or equl to ever number in S. For emple, if S is the set of numbers in the intervl (, 3), then u =, 4, 3., nd 3 re upper bounds for S nd l =,,.5, nd re lower bounds for S. Observe lso tht u = 3 is the smllest of ll upper bounds nd l = is the lrgest of ll lower bounds. The eistence of smllest upper bound nd lrgest lower bound for S is not ccidentl; it is consequence of the following iom iom (The Completeness Aiom) If nonempt set S of rel numbers hs n upper bound, then it hs smllest upper bound (clled the lest upper bound ), nd if nonempt set S of rel numbers hs lower bound, then it hs lrgest lower bound (clled the gretest lower bound). proof of theorem 9..3 () We will prove the result for incresing sequences, nd leve it for the reder to dpt the rgument to sequences tht re eventull incresing. Assume there eists number M such tht n M for n =,,...Then M is n upper bound for the set of terms in the sequence. B the Completeness Aiom there is lest upper bound for the terms; cll it L. Now let ɛ be n positive number. Since L is the lest upper bound for the terms, L ɛ is not n upper bound for the terms, which mens tht there is t lest one term N such tht N >L ɛ Moreover, since { n } is n incresing sequence, we must hve n N >L ɛ (6) when n N. But n cnnot eceed L since L is n upper bound for the terms. This observtion together with (6) tells us tht L n >L ɛ for n N, so ll terms from the Nth on re within ɛ units of L. This is ectl the requirement to hve lim n = L n + Finll, L M since M is n upper bound for the terms nd L is the lest upper bound. This proves prt (). (b) If there is no number M such tht n M for n =,,...,then no mtter how lrge we choose M, there is term N such tht nd, since the sequence is incresing, N >M n N >M when n N. Thus, the terms in the sequence become rbitrril lrge s n increses. Tht is, lim n =+ n + We omit the proof of Theorem 9..4 since it is similr to tht of 9..3.

201 9. Monotone Sequences 63 QUICK CHECK EXERCISES 9. (See pge 64 for nswers.). Clssif ech sequence s (I) incresing, (D) decresing, or (N) neither incresing nor decresing. {n} { n } { } { } 5 n n { ( ) n } n. Clssif ech sequence s (M) monotonic, (S) strictl monotonic, or (N) not monotonic. n {n + ( ) n } {n + ( ) n } {3n + ( ) n } 3. Since n/[(n + )] (n )/(n) = n n > the sequence {(n )/(n)} is strictl. 4. Since d d [( 8) ] > for > the sequence {(n 8) } is strictl. EXERCISE SET 9. 6 Use the difference n+ n to show tht the given sequence { n } is strictl incresing or strictl decresing. { } + {.. } + { } n + 3. n n= n n= n + n= { } n {n n } + n= 6. {n n } + n= 4n n= 7 Use the rtio n+ / n to show tht the given sequence { n } is strictl incresing or strictl decresing. { } n + { n } {ne n } + n + n= + n n= n= { n } + { n n } + { 5 n } +... (n)! n! n= n= (n ) 3 6 True Flse Determine whether the sttement is true or flse. Eplin our nswer. 3. If n+ n > for ll n, then the sequence { n } is strictl incresing. 4. A sequence { n } is monotone if n+ n = for ll n. 5. An bounded sequence converges. 6. If { n } is eventull incresing, then <. 7 Use differentition to show tht the given sequence is strictl incresing or strictl decresing. { } n + { } ln(n + ) n + n= n + n= 9. {tn n} + n=. {ne n } + n= 4 Show tht the given sequence is eventull strictl incresing or eventull strictl decresing. { }. {n 7n} + n + n=. n + n= { } n! {n 5 e n } + n= 3 n n= n= FOCUS ON CONCEPTS 5. Suppose tht { n } is monotone sequence such tht n for ll n. Must the sequence converge? If so, wht cn ou s bout the limit? 6. Suppose tht { n } is monotone sequence such tht n for ll n. Must the sequence converge? If so, wht cn ou s bout the limit? 7. Let { n } be the sequence defined recursivel b = nd n+ = + n for n. () List the first three terms of the sequence. (b) Show tht n < for n. (c) Show tht n+ n = ( n)( + n ) for n. (d) Use the results in prts (b) nd (c) to show tht { n } is strictl incresing sequence. [Hint: If nd re positive rel numbers such tht >, then it follows b fctoring tht >.] (e) Show tht { n } converges nd find its limit L. 8. Let { n } be the sequence defined recursivel b = nd n+ = [ n + (3/ n )] for n. () Show tht n 3 for n. [Hint: Wht is the minimum vlue of [ + (3 /)] for >?] (b) Show tht { n } is eventull decresing. [Hint: Emine n+ n or n+ / n nd use the result in prt ().] (c) Show tht { n } converges nd find its limit L. 9 3 The Beverton Holt model is used to describe chnges in popultion from one genertion to the net under certin ssumptions. If the popultion in genertion n is given b n, the Beverton Holt model predicts tht the popultion in the net genertion stisfies RK n n+ = K + (R ) n for some positive constnts R nd K with R>. These eercises eplore some properties of this popultion model.

202 64 Chpter 9 / Infinite Series 9. Let { n } be the sequence of popultion vlues defined recursivel b = 6, nd for n, n+ is given b the Beverton Holt model with R = nd K = 3. () List the first four terms of the sequence { n }. (b) If < n < 3, show tht < n+ < 3. Conclude tht < n < 3 for n. (c) Show tht { n } is incresing. (d) Show tht { n } converges nd find its limit L. 3. Let { n } be sequence of popultion vlues defined recursivel b the Beverton Holt model for which >K. Assume tht the constnts R nd K stisf R> nd K>. () If n >K, show tht n+ >K. Conclude tht n >K for ll n. (b) Show tht { n } is decresing. (c) Show tht { n } converges nd find its limit L. 3. The gol of this eercise is to estblish Formul (5), nmel, n lim n + n! = Let n = n /n! nd observe tht the cse where = is obvious, so we will focus on the cse where =. () Show tht n+ = n + n (b) Show tht the sequence { n } is eventull strictl decresing. (c) Show tht the sequence { n } converges. 3. () Compre pproprite res in the ccompning figure to deduce the following inequlities for n : n ln d<ln n! < n+ ln d (b) Use the result in prt () to show tht n n (n + )n+ <n! <, n > en e n (c) Use the Squeezing Theorem for Sequences (Theorem 9..5) nd the result in prt (b) to show tht = ln lim n + n n! n = e 3... n 3... n n + Figure E-3 = ln 33. Use the left inequlit in Eercise 3(b) to show tht n lim n!=+ n Writing Give n emple of n incresing sequence tht is not eventull strictl incresing. Wht cn ou conclude bout the terms of n such sequence? Eplin. 35. Writing Discuss the pproprite use of eventull for vrious properties of sequences. For emple, which is useful epression: eventull bounded or eventull monotone? QUICK CHECK ANSWERS 9.. I; D; N; I; N. N; M; S 3. ; incresing 4. 8; eventull; incresing 9.3 INFINITE SERIES The purpose of this section is to discuss sums tht contin infinitel mn terms. The most fmilir emples of such sums occur in the deciml representtions of rel numbers. For emple, when we write 3 in the deciml form = ,we men 3 = which suggests tht the deciml representtion of cn be viewed s sum of infinitel 3 mn rel numbers. SUMS OF INFINITE SERIES Our first objective is to define wht is ment b the sum of infinitel mn rel numbers. We begin with some terminolog.

203 9.3 Infinite Series definition An infinite series is n epression tht cn be written in the form u k = u + u + u 3 + +u k + k= The numbers u, u, u 3,...re clled the terms of the series..4 / ,.33,.333,... Figure 9.3. n Since it is impossible to dd infinitel mn numbers together directl, sums of infinite series re defined nd computed b n indirect limiting process. To motivte the bsic ide, consider the deciml () This cn be viewed s the infinite series or, equivlentl, () 3 4 Since () is the deciml epnsion of, n resonble definition for the sum of n infinite 3 series should ield for the sum of (). To obtin such definition, consider the following 3 sequence of (finite) sums: s = 3 =.3 s = =.33 s 3 = =.333 s 4 = = The sequence of numbers s, s, s 3, s 4,... (Figure 9.3.) cn be viewed s succession of pproimtions to the sum of the infinite series, which we wnt to be. As we 3 progress through the sequence, more nd more terms of the infinite series re used, nd the pproimtions get better nd better, suggesting tht the desired sum of might be the limit 3 of this sequence of pproimtions. To see tht this is so, we must clculte the limit of the generl term in the sequence of pproimtions, nmel, s n = (3) n The problem of clculting ( 3 lim s n = lim n + n ) n is complicted b the fct tht both the lst term nd the number of terms in the sum chnge with n. It is best to rewrite such limits in closed form in which the number of terms does not vr, if possible. (See the discussion of closed form nd open form following Emple in Section 5.4.) To do this, we multipl both sides of (3) b to obtin s n = (4) n n+

204 66 Chpter 9 / Infinite Series nd then subtrct (4) from (3) to obtin s n s n = s n = 3 s n = 3 Since / n sn +, it follows tht which we denote b writing lim n + s n = ( n ( n lim n + 3 ) n+ ) ( ) = n 3 3 = n Motivted b the preceding emple, we re now red to define the generl concept of the sum of n infinite series u + u + u 3 + +u k + We begin with some terminolog: Let s n denote the sum of the initil terms of the series, up to nd including the term with inde n. Thus, s = u s = u + u s 3 = u + u + u 3. s n = u + u + u 3 + +u n = The number s n is clled the nth prtil sum of the series nd the sequence {s n } + n= is clled the sequence of prtil sums. As n increses, the prtil sum s n = u + u + +u n includes more nd more terms of the series. Thus, if s n tends towrd limit s n +, it is resonble to view this limit s the sum of ll the terms in the series. This suggests the following definition. n k= u k WARNING In everd lnguge the words sequence nd series re often used interchngebl. However, in mthemtics there is distinction between these two words sequence is succession wheres series is sum. Itis essentil tht ou keep this distinction in mind definition Let {s n } be the sequence of prtil sums of the series u + u + u 3 + +u k + If the sequence {s n } converges to limit S, then the series is sid to converge to S, nd S is clled the sum of the series. We denote this b writing S = k= If the sequence of prtil sums diverges, then the series is sid to diverge. A divergent series hs no sum. u k Emple Determine whether the series converges or diverges. If it converges, find the sum.

205 ,,,,,,... Figure 9.3. n 9.3 Infinite Series 67 Solution. It is tempting to conclude tht the sum of the series is zero b rguing tht the positive nd negtive terms cncel one nother. However, this is not correct; the problem is tht lgebric opertions tht hold for finite sums do not crr over to infinite series in ll cses. Lter, we will discuss conditions under which fmilir lgebric opertions cn be pplied to infinite series, but for this emple we turn directl to Definition The prtil sums re s = s = = s 3 = + = s 4 = + = nd so forth. Thus, the sequence of prtil sums is,,,,,,... (Figure 9.3.). Since this is divergent sequence, the given series diverges nd consequentl hs no sum. GEOMETRIC SERIES In mn importnt series, ech term is obtined b multipling the preceding term b some fied constnt. Thus, if the initil term of the series is nd ech term is obtined b multipling the preceding term b r, then the series hs the form r k = + r + r + r 3 + +r k + ( = ) (5) k= Such series re clled geometric series, nd the number r is clled the rtio for the series. Here re some emples: k + =,r = k = 3,r = ( )k+ + k =,r = =,r = + + +( ) k+ + =,r = Sometimes it is desirble to strt the inde of summtion of n infinite series t k = rther thn k =, in which cse we would cll u the zeroth term nd s = u the zeroth prtil sum. One cn prove tht chnging the strting vlue for the inde of summtion of n infinite series hs no effect on the convergence, the divergence, or the sum. If we hd strted the inde t k = in (5), then the series would be epressed s r k k= Since this epression is more complicted thn (5), we strted the inde t k = k + =,r = The following theorem is the fundmentl result on convergence of geometric series theorem A geometric series r k = + r + r + +r k + ( = ) k= converges if r < nd diverges if r. If the series converges, then the sum is proof r k = r k= Let us tret the cse r = first. If r =, then the series is

206 68 Chpter 9 / Infinite Series so the nth prtil sum is s n = (n + ) nd lim s n = lim (n + ) =± n + n + (the sign depending on whether is positive or negtive). This proves divergence. If r =, the series is + + so the sequence of prtil sums is,,,,,,... which diverges. Now let us consider the cse where r =. The nth prtil sum of the series is Multipling both sides of (6) b r ields nd subtrcting (7) from (6) gives or s n = + r + r + +r n (6) rs n = r + r + +r n + r n+ (7) s n rs n = r n+ ( r)s n = r n+ (8) Since r = in the cse we re considering, this cn be rewritten s s n = rn+ r = r ( rn+ ) (9) Note tht (6) is n open form for s n, while (9) is closed form for s n. In generl, one needs closed form to clculte the limit. If r <, then r n+ goes to s n + (cn ou see wh?), so {s n } converges. From (9) lim s n = n + r If r >, then either r>orr<. In the cse r>, r n+ increses without bound s n +, nd in the cse r<, r n+ oscilltes between positive nd negtive vlues tht grow in mgnitude, so {s n } diverges in both cses. /3 6 5 Emple sum. In ech prt, determine whether the series converges, nd if so find its 5 () (b) 3 k 5 k 4 k k= k= Prtil sums for k= 4 k Figure n Solution (). This is geometric series with = 5 nd r = 4. Since r = <, the 4 series converges nd the sum is r = 5 = 3 4 (Figure 9.3.3). Solution (b). This is geometric series in conceled form, since we cn rewrite it s 3 k 5 k 9 k ( ) 9 k = 5 = 9 k 5 k= Since r = 9 >, the series diverges. 5 k= k=

207 9.3 Infinite Series 69 TECHNOLOGY MASTERY Computer lgebr sstems hve commnds for finding sums of convergent series. If ou hve CAS, use it to compute the sums in Emples nd 3. Emple 3 Solution. Find the rtionl number represented b the repeting deciml We cn write = so the given deciml is the sum of geometric series with =.784 nd r =.. Thus, = r =.784. = = Emple 4 In ech prt, find ll vlues of for which the series converges, nd find the sum of the series for those vlues of. () k (b) ( )k + + k + 8 k Solution (). k= The epnded form of the series is k = k + k= The series is geometric series with = nd r =, so it converges if < nd diverges otherwise. When the series converges its sum is k = k= Solution (b). This is geometric series with = 3 nd r = /. It converges if / <, or equivlentl, when <. When the series converges its sum is k= ( 3 ) k = 3 ( ) = 6 + TELESCOPING SUMS Emple 5 Determine whether the series k(k + ) = k= converges or diverges. If it converges, find the sum. Solution. The nth prtil sum of the series is n s n = k(k + ) = n(n + ) k= We will begin b rewriting s n in closed form. This cn be ccomplished b using the method of prtil frctions to obtin (verif) k(k + ) = k k +

208 6 Chpter 9 / Infinite Series The sum in () is n emple of telescoping sum. The nme is derived from the fct tht in simplifing the sum, one term in ech prentheticl epression cncels one term in the net prentheticl epression, until the entire sum collpses (like folding telescope) into just two terms. n () (b) {s n } {s n} Prtil sums for the hrmonic series Figure Courtes Lill Librr, Indin Universit This is proof of the divergence of the hrmonic series, s it ppered in n ppendi of Jkob Bernoulli s posthumous publiction, Ars Conjectndi, which ppered in 73. n from which we obtin the sum n ( s n = k ) k + k= ( = ) ( + 3 ( = + + ) + Thus, = n + k= k(k + ) = ) ( + ( lim s n = n ) ( + + ) + + n ) n + ) n + ( n + n ( lim ) = n + n + HARMONIC SERIES One of the most importnt of ll diverging series is the hrmonic series, k= k = which rises in connection with the overtones produced b vibrting musicl string. It is not immeditel evident tht this series diverges. However, the divergence will become pprent when we emine the prtil sums in detil. Becuse the terms in the series re ll positive, the prtil sums s =, s = +, s 3 = + + 3, s 4 = ,... form strictl incresing sequence s <s <s 3 < <s n < (Figure 9.3.4). Thus, b Theorem 9..3 we cn prove divergence b demonstrting tht there is no constnt M tht is greter thn or equl to ever prtil sum. To this end, we will consider some selected prtil sums, nmel, s, s 4, s 8, s 6, s 3,...Note tht the subscripts re successive powers of, so tht these re the prtil sums of the form s n (Figure 9.3.4b). These prtil sums stisf the inequlities s = + > + = s 4 = s >s + ( 4 + 4) = s + > 3 s 8 = s >s 4 + ( ) = s4 + > 4 s 6 = s >s 8 + ( ) = s8 + > 5 s n > n + If M is n constnt, we cn find positive integer n such tht (n + )/ >M. But for this n s n > n + >M so tht no constnt M is greter thn or equl to ever prtil sum of the hrmonic series. This proves divergence. This divergence proof, which predtes the discover of clculus, is due to French bishop nd techer, Nicole Oresme (33 38). This series eventull ttrcted the interest of Johnn nd Jkob Bernoulli (p. 7) nd led them to begin thinking bout the generl concept of convergence, which ws new ide t tht time. ()

209 9.3 Infinite Series 6 QUICK CHECK EXERCISES 9.3 (See pge 63 for nswers.). In mthemtics, the terms sequence nd series hve different menings: is succession, wheres is sum.. Consider the series k k= If {s n } is the sequence of prtil sums for this series, then s =, s =, s 3 =, s 4 =, nd s n =. 3. Wht does it men to s tht series u k converges? 4. A geometric series is series of the form k= This series converges to if. This series diverges if. 5. The hrmonic series hs the form k= Does the hrmonic series converge or diverge? EXERCISE SET 9.3 C CAS In ech prt, find ect vlues for the first four prtil sums, find closed form for the nth prtil sum, nd determine whether the series converges b clculting the limit of the nth prtil sum. If the series converges, then stte its sum.. () k (b) (c). () + + k ( ) k (b) 4 k (c) 4 k= k= (k + )(k + ) + ( k + 3 ) k Determine whether the series converges, nd if so find its sum. ( 3. 3 ) k ( ) k k= ( ) k k k= k= k= k=3 k= (k + )(k + 3) 9k + 3k k k+ 7 k 4. k= k= k= k= k= ( 3 ) k+ ( ) k k+ k ( e k π) k=5 5 3k 7 k k= 5. Mtch series from one of Eercises 3, 5, 7, or 9 with the grph of its sequence of prtil sums. () (b) n. n (c) n (d) Mtch series from one of Eercises 4, 6, 8, or with the grph of its sequence of prtil sums. () (b) n n (c) n (d) n n

210 6 Chpter 9 / Infinite Series 7 True Flse Determine whether the sttement is true or flse. Eplin our nswer. 7. An infinite series converges if its sequence of terms converges. 8. The geometric series + r + r + +r n + converges provided r <. 9. The hrmonic series diverges.. An infinite series converges if its sequence of prtil sums is bounded nd monotone. 4 Epress the repeting deciml s frction Recll tht terminting deciml is deciml whose digits re ll from some point on (.5 =.5...,for emple). Show tht deciml of the form.... n , where n = 9, cn be epressed s terminting deciml. FOCUS ON CONCEPTS 6. The gret Swiss mthemticin Leonhrd Euler (biogrph on p. 3) sometimes reched incorrect conclusions in his pioneering work on infinite series. For emple, Euler deduced tht = + + nd = b substituting = nd = in the formul = Wht ws the problem with his resoning? 7. A bll is dropped from height of m. Ech time it strikes the ground it bounces verticll to height tht is 3 of the preceding height. Find the totl distnce the 4 bll will trvel if it is ssumed to bounce infinitel often. 8. The ccompning figure shows n infinite stircse constructed from cubes. Find the totl volume of the stircse, given tht the lrgest cube hs side of length nd ech successive cube hs side whose length is hlf tht of the preceding cube.... Figure E-8 9. In ech prt, find closed form for the nth prtil sum of the series, nd determine whether the series converges. If so, find its sum. () ln + ln 3 + ln ln k k + + ( (b) ln ) ( + ln ) ( + ln k= ) + ( ) + ln + (k + ) 3. Use geometric series to show tht () ( ) k k = if << + k= (b) ( 3) k = if <<4 4 k= (c) ( ) k k = if << In ech prt, find ll vlues of for which the series converges, nd find the sum of the series for those vlues of. () (b) (c) e + e + e 3 + e 4 + e Show tht for ll rel vlues of sin sin + 4 sin3 8 sin4 + = sin + sin 33. Let be n rel number, nd let { n } be the sequence defined recursivel b n+ = ( n + ) Mke conjecture bout the limit of the sequence, nd confirm our conjecture b epressing n in terms of nd tking the limit. k + k 34. Show: =. k + k k= ( 35. Show: k ) = 3 k +. k= 36. Show: = Show: =. 38. In his Tretise on the Configurtions of Qulities nd Motions (written in the 35s), the French Bishop of Lisieu, Nicole Oresme, used geometric method to find the sum of the series k= k k = In prt () of the ccompning figure, ech term in the series is represented b the re of rectngle, nd in

211 9.4 Convergence Tests 63 prt (b) the configurtion in prt () hs been divided into rectngles with res A, A, A 3,... Find the sum A + A + A 3 +. Source: This problem is bsed on Trisection of n Angle in n Infinite Number of Steps b Eric Kincnnon, which ppered in The College Mthemtics Journl, Vol., No. 5, November 99. R Figure E-38 () Not to scle A (b) A 3 A 39. As shown in the ccompning figure, suppose tht n ngle θ is bisected using strightedge nd compss to produce r R, then the ngle between R nd the initil side is bisected to produce r R. Therefter, rs R 3, R 4, R 5,... re constructed in succession b bisecting the ngle between the preceding two rs. Show tht the sequence of ngles tht these rs mke with the initil side hs limit of θ/3. C 4. k= u R 3 R4 R Initil side k= Figure E-39 In ech prt, use CAS to find the sum of the series if it converges, nd then confirm the result b hnd clcultion. () ( ) k+ k 3 k 3 3k (b) (c) 5 k 4k 4. Writing Discuss the similrities nd differences between wht it mens for sequence to converge nd wht it mens for series to converge. 4. Writing Red bout Zeno s dichotom prdo in n pproprite reference work nd relte the prdo in setting tht is fmilir to ou. Discuss connection between the prdo nd geometric series. k= QUICK CHECK ANSWERS 9.3. sequence; series. ; 3 4 ; 7 8 ; 5 6 ; 3. The sequence of prtil sums converges. n 4. r k ( = ); r ; r < ; r 5. k ; diverge 9.4 CONVERGENCE TESTS In the lst section we showed how to find the sum of series b finding closed form for the nth prtil sum nd tking its limit. However, it is reltivel rre tht one cn find closed form for the nth prtil sum of series, so lterntive methods re needed for finding the sum of series. One possibilit is to prove tht the series converges, nd then to pproimte the sum b prtil sum with sufficientl mn terms to chieve the desired degree of ccurc. In this section we will develop vrious tests tht cn be used to determine whether given series converges or diverges. THE DIVERGENCE TEST In stting generl results bout convergence or divergence of series, it is convenient to use the nottion u k s generic nottion for series, thus voiding the issue of whether the sum begins with k = ork = or some other vlue. Indeed, we will see shortl tht the strting inde vlue is irrelevnt to the issue of convergence. The kth term in n infinite series u k is clled the generl term of the series. The following theorem estblishes

212 64 Chpter 9 / Infinite Series reltionship between the limit of the generl term nd the convergence properties of series theorem (The Divergence Test) () (b) If lim u k =, then the series u k diverges. k + If lim u k =, then the series u k m either converge or diverge. k + proof () To prove this result, it suffices to show tht if the series converges, then lim k + u k = (wh?). We will prove this lterntive form of (). Let us ssume tht the series converges. The generl term u k cn be written s u k = s k s k () where s k is the sum of the terms through u k nd s k is the sum of the terms through u k.if S denotes the sum of the series, then lim k + s k = S, nd since (k ) + s k +, we lso hve lim k + s k = S. Thus, from () lim k + u k = lim k + (s k s k ) = S S = proof (b) To prove this result, it suffices to produce both convergent series nd divergent series for which lim k + u k =. The following series both hve this propert: nd + k k + The first is convergent geometric series nd the second is the divergent hrmonic series. WARNING The converse of Theorem 9.4. is flse; tht is, showing tht lim u k = k + does not prove tht u k converges, since this propert m hold for divergent s well s convergent series. This is illustrted in the proof of prt (b) of Theorem The lterntive form of prt () given in the preceding proof is sufficientl importnt tht we stte it seprtel for future reference theorem If the series u k converges, then lim k + u k =. Emple diverges since The series k= k k + = lim k + k k + = lim k + k k /k = = ALGEBRAIC PROPERTIES OF INFINITE SERIES For brevit, the proof of the following result is omitted.

213 9.4 Convergence Tests 65 See Eercises 7 nd 8 for n eplortion of wht hppens when u k or vk diverge theorem () (b) If u k nd v k re convergent series, then (u k + v k ) nd (u k v k ) re convergent series nd the sums of these series re relted b (u k + v k ) = k= (u k v k ) = k= u k + k= u k If c is nonzero constnt, then the series u k nd cu k both converge or both diverge. In the cse of convergence, the sums re relted b k= cu k = c k= k= u k k= k= v k v k WARNING Do not red too much into prt (c) of Theorem Although convergence is not ffected when finitel mn terms re deleted from the beginning of convergent series, the sum of the series is chnged b the removl of those terms. (c) Convergence or divergence is unffected b deleting finite number of terms from series; in prticulr, for n positive integer K, the series u k = u + u + u 3 + k= u k = u K + u K+ + u K+ + k=k both converge or both diverge. Emple Find the sum of the series ( 3 4 ) k 5 k k= Solution. The series k= 3 4 k = is convergent geometric series ( = 3 4,r = 4), nd the series k= k= 5 k = is lso convergent geometric series ( =,r = 5). Thus, from Theorems 9.4.3() nd the given series converges nd ( 3 4 ) 3 = k 5 k 4 k 5 k k= k= 3 4 = 4 5 = 3

214 66 Chpter 9 / Infinite Series Emple 3 Determine whether the following series converge or diverge. () k= 5 k = k + (b) k = k= Solution. The first series is constnt times the divergent hrmonic series, nd hence diverges b prt (b) of Theorem The second series results b deleting the first nine terms from the divergent hrmonic series, nd hence diverges b prt (c) of Theorem THE INTEGRAL TEST The epressions k= k nd + d u u u3 u4 = f() re relted in tht the integrnd in the improper integrl results when the inde k in the generl term of the series is replced b nd the limits of summtion in the series re replced b the corresponding limits of integrtion. The following theorem shows tht there is reltionship between the convergence of the series nd the integrl. u n n + () u u3 u4 = f () theorem (The Integrl Test) Let u k be series with positive terms. If f is function tht is decresing nd continuous on n intervl [,+ ) nd such tht u k = f(k) for ll k, then + nd f()d both converge or both diverge. u k k= u n n (b) Figure 9.4. The proof of the integrl test is deferred to the end of this section. However, the gist of the proof is cptured in Figure 9.4.: if the integrl diverges, then so does the series (Figure 9.4.), nd if the integrl converges, then so does the series (Figure 9.4.b). Emple 4 Show tht the integrl test pplies, nd use the integrl test to determine whether the following series converge or diverge. () (b) k k k= Solution (). We lred know tht this is the divergent hrmonic series, so the integrl test will simpl illustrte nother w of estblishing the divergence. Note first tht the series hs positive terms, so the integrl test is pplicble. If we replce k b in the generl term /k, we obtin the function f() = /, which is decresing nd continuous for (s required to ppl the integrl test with = ). Since + d = lim b + b k= d = lim [ln b ln ] =+ b + the integrl diverges nd consequentl so does the series.

215 9.4 Convergence Tests 67 WARNING In prt (b) of Emple 4, do not erroneousl conclude tht the sum of the series is becuse the vlue of the corresponding integrl is. You cn see tht this is not so since the sum of the first two terms lone eceeds. Lter, we will see tht the sum of the series is ctull π /6. Solution (b). Note first tht the series hs positive terms, so the integrl test is pplicble. If we replce k b in the generl term /k, we obtin the function f() = /, which is decresing nd continuous for. Since + b [ d d = lim b + = lim ] b [ = lim ] = b + b + b the integrl converges nd consequentl the series converges b the integrl test with =. p-series The series in Emple 4 re specil cses of clss of series clled p-series or hperhrmonic series. Ap-series is n infinite series of the form k = + p + p p k + p k= where p>. Emples of p-series re k = k + k= p = k = k + k= p = = k 3 k p = k= The following theorem tells when p-series converges theorem (Convergence of p-series) k = + p + p p k + p k= converges if p> nd diverges if <p. proof + To estblish this result when p =, we will use the integrl test. d = lim p b + b p p d = lim b + p ] b = lim b + [ b p p p Assume first tht p>. Then p<, so b p sb +. Thus, the integrl converges [its vlue is /( p)] nd consequentl the series lso converges. Now ssume tht <p<. It follows tht p> nd b p + s b +, so the integrl nd the series diverge. The cse p = is the hrmonic series, which ws previousl shown to diverge. ] Emple k + diverges since it is p-series with p = 3 <.

216 68 Chpter 9 / Infinite Series PROOF OF THE INTEGRAL TEST Before we cn prove the integrl test, we need bsic result bout convergence of series with nonnegtive terms. If u + u + u 3 + +u k + is such series, then its sequence of prtil sums is incresing, tht is, s s s 3 s n Thus, from Theorem 9..3 the sequence of prtil sums converges to limit S if nd onl if it hs some upper bound M, in which cse S M. If no upper bound eists, then the sequence of prtil sums diverges. Since convergence of the sequence of prtil sums corresponds to convergence of the series, we hve the following theorem theorem If constnt M such tht uk is series with nonnegtive terms, nd if there is s n = u + u + +u n M for ever n, then the series converges nd the sum S stisfies S M. If no such M eists, then the series diverges. In words, this theorem implies tht series with nonnegtive terms converges if nd onl if its sequence of prtil sums is bounded bove. u u u3 u4 = f() u n n + () u u3 u4 = f () u n n (b) Figure 9.4. proof of theorem We need onl show tht the series converges when the integrl converges nd tht the series diverges when the integrl diverges. For simplicit, we will limit the proof to the cse where =. Assume tht f() stisfies the hpotheses of the theorem for. Since f() = u,f() = u,...,f(n)= u n,... the vlues of u, u,...,u n,...cn be interpreted s the res of the rectngles shown in Figure The following inequlities result b compring the res under the curve = f() to the res of the rectngles in Figure 9.4. for n>: n+ f()d < u + u + +u n = s n s n u = u + u 3 + +u n < These inequlities cn be combined s n+ n f()d < s n <u + f()d n Figure 9.4. Figure 9.4.b f()d () If the integrl + f()d converges to finite vlue L, then from the right-hnd inequlit in () n + s n <u + f()d < u + f()d = u + L Thus, ech prtil sum is less thn the finite constnt u + L, nd the series converges b Theorem On the other hnd, if the integrl + f()d diverges, then lim n + n+ f()d =+ so tht from the left-hnd inequlit in (), s n + s n +. This implies tht the series lso diverges.

217 9.4 Convergence Tests 69 QUICK CHECK EXERCISES 9.4 (See pge 63 for nswers.). The divergence test ss tht if =, then the series uk diverges.. Given tht = 3, it follows tht k = k= k =, k= nd nd b k = 5 k= ( k + b k ) = k= 3. Since + (/ )d =+, the test pplied to the series k= shows tht this series. 4. A p-series is series of the form k= This series converges if.. This series diverges if EXERCISE SET 9.4 Grphing Utilit C CAS. Use ( Theorem to find the sum of ech series. () + ) ( ) ( ) + k 4 k ( ) (b) 5 k k(k + ) k=. Use Theorem to find the sum of ech series. [ () k 7 ] ] (b) [7 k 3 k+ k+ k 5 k k= 3 4 For ech given p-series, identif p nd determine whether the series converges. 3. () (b) (c) k (d) k /3 k 3 k 4. () k= k 4/3 k= (b) k= k= 4 k (c) k= k= k= 3 k 5 (d) k= k= k π 5 6 Appl the divergence test nd stte wht it tells ou bout the series. k + k + 3 ( 5. () (b) + ) k k + k k= k= (c) cos kπ (d) k! 6. () (c) k= k= k= k e k k (b) (d) k= ln k k k Confirm tht the integrl test is pplicble nd use it to determine whether the series converges. k= k= 7. () 8. () k= k= k= 5k + (b) k + k (b) + 9k (4 + k) 3/ k= k= 9 4 Determine whether the series converges k + 6 5k k k= k= k= k e 3. k ln(k + ) 6. k + k ( ) k sin k k= k= k= 3 k 4. ke k 7. k= k= tn k + k.. 7k. 4. k=5 k= k e k3 k= sech k k= k= ln k k ( + k k=3 k= k= k + ) k 5 6 Use the integrl test to investigte the reltionship between the vlue of p nd the convergence of the series k(ln k) p k(ln k)[ln(ln k)] p FOCUS ON CONCEPTS k=3 7. Suppose tht the series u k converges nd the series v k diverges. Show tht the series (u k + v k ) nd (u k v k ) both diverge. [Hint: Assume tht (uk + v k ) converges nd use Theorem to obtin contrdiction.]

218 63 Chpter 9 / Infinite Series 8. Find emples to show tht if the series u k nd vk both diverge, then the series (u k + v k ) nd (uk v k ) m either converge or diverge. = f () C 9 3 Use the results of Eercises 7 nd 8, if needed, to determine whether ech series converges or diverges. [ ( ) ] k 9. () + [ (b) 3 k 3k + ] k 3/ 3. () k= [ k(ln k) ] k k= (b) k= k= [ ke k + ] k ln k 3 34 True Flse Determine whether the sttement is true or flse. Eplin our nswer. 3. If u k converges to L, then (/u k ) converges to /L. 3. If cu k diverges for some constnt c, then u k must diverge. 33. The integrl test cn be used to prove tht series diverges. 34. The series is p-series. pk 35. k= Use CAS to confirm tht k= k= k = π 6 nd k= k 4 = π4 9 nd then use these results in ech prt to find the sum of the series. 3k () (b) (c) k 4 k (k ) 4 k= Eercise 36 will show how prtil sum cn be used to obtin upper nd lower bounds on the sum of series when the hpotheses of the integrl test re stisfied. This result will be needed in Eercises () Let k= u k be convergent series with positive terms, nd let f be function tht is decresing nd continuous on [n, + ) nd such tht u k = f(k)for k n. Use n re rgument nd the ccompning figure to show tht + n+ f()d < k=n+ u k < + n k= f()d (b) Show tht if S is the sum of the series k= u k nd s n is the nth prtil sum, then s n + + n+ f()d<s<s n + + n f()d... n n + = f ()... n n + u n+un+... u n+ u n+un () It ws stted in Eercise 35 tht k= k = π 6 Figure E-36 Show tht if s n is the nth prtil sum of this series, then s n + n + < π 6 <s n + n (b) Clculte s 3 ectl, nd then use the result in prt () to show tht 9 8 < π 6 < 6 36 (c) Use clculting utilit to confirm tht the inequlities in prt (b) re correct. (d) Find upper nd lower bounds on the error tht results if the sum of the series is pproimted b the th prtil sum. 38. In ech prt, find upper nd lower bounds on the error tht results if the sum of the series is pproimted b the th prtil sum. () k= (k + ) (b) k= k It ws stted in Eercise 35 tht k = π4 4 9 k= (c) k= k e k () Let s n be the nth prtil sum of the series bove. Show tht s n + 3(n + ) < π4 3 9 <s n + 3n 3 (b) We cn use prtil sum of the series to pproimte π 4 /9 to three deciml-plce ccurc b cpturing the

219 sum of the series in n intervl of length. (or less). Find the smllest vlue of n such tht the intervl contining π 4 /9 in prt () hs length of. or less. (c) Approimte π 4 /9 to three deciml plces using the midpoint of n intervl of width t most. tht contins the sum of the series. Use clculting utilit to confirm tht our nswer is within.5 of π 4 /9. 4. We showed in Section 9.3 tht the hrmonic series k= /k diverges. Our objective in this problem is to demonstrte tht lthough the prtil sums of this series pproch +, the increse etremel slowl. () Use inequlit () to show tht for n ln(n + ) <s n < + ln n (b) Use the inequlities in prt () to find upper nd lower bounds on the sum of the first million terms in the series. C 9.5 The Comprison, Rtio, nd Root Tests 63 (c) Show tht the sum of the first billion terms in the series is less thn. (d) Find vlue of n so tht the sum of the first n terms is greter thn. 4. Use grphing utilit to confirm tht the integrl test pplies to the series k= k e k, nd then determine whether the series converges. 4. () Show tht the hpotheses of the integrl test re stisfied b the series k= /(k 3 + ). (b) Use CAS nd the integrl test to confirm tht the series converges. (c) Construct tble of prtil sums for n =,, 3,...,, showing t lest si deciml plces. (d) Bsed on our tble, mke conjecture bout the sum of the series to three deciml-plce ccurc. (e) Use prt (b) of Eercise 36 to check our conjecture. QUICK CHECK ANSWERS 9.4. lim k + u k. ; 7 3. integrl; k ; diverges 4. ; p>; <p kp 9.5 THE COMPARISON, RATIO, AND ROOT TESTS In this section we will develop some more bsic convergence tests for series with nonnegtive terms. Lter, we will use some of these tests to stud the convergence of Tlor series. THE COMPARISON TEST We will begin with test tht is useful in its own right nd is lso the building block for other importnt convergence tests. The underling ide of this test is to use the known convergence or divergence of series to deduce the convergence or divergence of nother series. It is not essentil in Theorem 9.5. tht the condition k b k hold for ll k, s stted; the conclusions of the theorem remin true if this condition is eventull true theorem (The Comprison Test) Let k= k nd k= b k be series with nonnegtive terms nd suppose tht () (b) b, b, 3 b 3,..., k b k,... If the bigger series b k converges, then the smller series k lso converges. If the smller series k diverges, then the bigger series b k lso diverges. We hve left the proof of this theorem for the eercises; however, it is es to visulize wh the theorem is true b interpreting the terms in the series s res of rectngles

220 63 Chpter 9 / Infinite Series (Figure 9.5.). The comprison test sttes tht if the totl re b k is finite, then the totl re k must lso be finite; nd if the totl re k is infinite, then the totl re b k must lso be infinite. b b b 3 b 4 b 5 3 b k k k... For ech rectngle, k denotes the re of the blue portion nd b k denotes the combined re of the white nd blue portions. Figure 9.5. USING THE COMPARISON TEST There re two steps required for using the comprison test to determine whether series uk with positive terms converges: Step. Guess t whether the series u k converges or diverges. Step. Find series tht proves the guess to be correct. Tht is, if we guess tht u k diverges, we must find divergent series whose terms re smller thn the corresponding terms of u k, nd if we guess tht u k converges, we must find convergent series whose terms re bigger thn the corresponding terms of u k. In most cses, the series u k being considered will hve its generl term u k epressed s frction. To help with the guessing process in the first step, we hve formulted two principles tht re bsed on the form of the denomintor for u k. These principles sometimes suggest whether series is likel to converge or diverge. We hve clled these informl principles becuse the re not intended s forml theorems. In fct, we will not gurntee tht the lws work. However, the work often enough to be useful informl principle Constnt terms in the denomintor of u k cn usull be deleted without ffecting the convergence or divergence of the series informl principle If polnomil in k ppers s fctor in the numertor or denomintor of u k, ll but the leding term in the polnomil cn usull be discrded without ffecting the convergence or divergence of the series. Emple Use the comprison test to determine whether the following series converge or diverge. () (b) k k + k k= Solution (). According to Principle 9.5., we should be ble to drop the constnt in the denomintor without ffecting the convergence or divergence. Thus, the given series is likel to behve like () k k= which is divergent p-series ( p = ). Thus, we will guess tht the given series diverges nd tr to prove this b finding divergent series tht is smller thn the given series. However, series () does the trick since > for k =,,... k k Thus, we hve proved tht the given series diverges. k=

221 k= 9.5 The Comprison, Rtio, nd Root Tests 633 Solution (b). According to Principle 9.5.3, we should be ble to discrd ll but the leding term in the polnomil without ffecting the convergence or divergence. Thus, the given series is likel to behve like k = () k which converges since it is constnt times convergent p-series (p = ). Thus, we will guess tht the given series converges nd tr to prove this b finding convergent series tht is bigger thn the given series. However, series () does the trick since k + k < for k =,,... k Thus, we hve proved tht the given series converges. k= THELIMITCOMPARISONTEST In the lst emple, Principles 9.5. nd provided the guess bout convergence or divergence s well s the series needed to ppl the comprison test. Unfortuntel, it is not lws so strightforwrd to find the series required for comprison, so we will now consider n lterntive to the comprison test tht is usull esier to ppl. The proof is given in Appendi D theorem (The Limit Comprison Test) Let k nd b k be series with positive terms nd suppose tht k ρ = lim k + b k If ρ is finite nd ρ>, then the series both converge or both diverge. The cses where ρ = orρ =+ re discussed in the eercises (Eercise 56). To use the limit comprison test we must gin first guess t the convergence or divergence of k nd then find series b k tht supports our guess. The following emple illustrtes this principle. Emple Use the limit comprison test to determine whether the following series converge or diverge. 3k 3 k + 4 () (b) (c) k + k + k k 7 k 3 + k= k= Solution (). As in Emple, Principle 9.5. suggests tht the series is likel to behve like the divergent p-series (). To prove tht the given series diverges, we will ppl the limit comprison test with k = nd b k = k + k k= We obtin k ρ = lim = lim k + b k k + k k + = lim k + + = k Since ρ is finite nd positive, it follows from Theorem tht the given series diverges.

222 634 Chpter 9 / Infinite Series Solution (b). As in Emple, Principle suggests tht the series is likel to behve like the convergent series (). To prove tht the given series converges, we will ppl the limit comprison test with k = k + k nd b k = k We obtin k k ρ = lim = lim k + b k k + k + k = lim k + + = k Since ρ is finite nd positive, it follows from Theorem tht the given series converges, which grees with the conclusion reched in Emple using the comprison test. Solution (c). From Principle 9.5.3, the series is likel to behve like 3k 3 k = 3 (3) 7 k 4 k= which converges since it is constnt times convergent p-series. Thus, the given series is likel to converge. To prove this, we will ppl the limit comprison test to series (3) nd the given series. We obtin ρ = lim k + 3k 3 k + 4 k 7 k k 4 k= 3k 7 k 6 + 4k 4 = lim k + 3k 7 3k = Since ρ is finite nd nonzero, it follows from Theorem tht the given series converges, since (3) converges. THE RATIO TEST The comprison test nd the limit comprison test hinge on first mking guess bout convergence nd then finding n pproprite series for comprison, both of which cn be difficult tsks in cses where Principles 9.5. nd cnnot be pplied. In such cses the net test cn often be used, since it works eclusivel with the terms of the given series it requires neither n initil guess bout convergence nor the discover of series for comprison. Its proof is given in Appendi D theorem (The Rtio Test) Let u k be series with positive terms nd suppose tht u k+ ρ = lim k + u k () If ρ<, the series converges. (b) (c) If ρ> or ρ =+, the series diverges. If ρ =, the series m converge or diverge, so tht nother test must be tried. Emple 3 Ech of the following series hs positive terms, so the rtio test pplies. In ech prt, use the rtio test to determine whether the following series converge or diverge. k k k (k)! () (b) (c) (d) (e) k! k k! 4 k k k= k= k= k=3 k=

223 Solution (). The series converges, since u k+ /(k + )! ρ = lim = lim = lim k + u k k + /k! k + Solution (b). Solution (c). The series converges, since u k+ ρ = lim = lim k + u k k + The series diverges, since 9.5 The Comprison, Rtio, nd Root Tests 635 k + k k+ k = k! (k + )! = lim k + k + = < lim k + = k + k < u k+ (k + ) k+ ρ = lim = lim k! k + u k k + (k + )! k = lim (k + ) k = lim k k + k k k + ( + k) k = e> See Formul (7) of Section.3 Solution (d). u k+ ρ = lim k + = lim k + The series diverges, since [(k + )]! = lim u k k + 4 k+ ( (k + )(k + )(k)! (k)! 4 4 k (k)! = lim k + ) ( (k + )! (k)! ) 4 = 4 lim (k + )(k + ) =+ k + Solution (e). The rtio test is of no help since u k+ ρ = lim = lim k + u k k + (k + ) k k = lim k + k + = However, the integrl test proves tht the series diverges since + d = lim b + b d = lim ln( ) b + ] b =+ Both the comprison test nd the limit comprison test would lso hve worked here (verif). THE ROOT TEST In cses where it is difficult or inconvenient to find the limit required for the rtio test, the net test is sometimes useful. Since its proof is similr to the proof of the rtio test, we will omit it theorem (The Root Test) Let u k be series with positive terms nd suppose tht ρ = lim k uk = lim (u k) /k k + k + () If ρ<, the series converges. (b) (c) If ρ> or ρ =+, the series diverges. If ρ =, the series m converge or diverge, so tht nother test must be tried. Emple 4 diverge. Use the root test to determine whether the following series converge or ( ) 4k 5 k () (b) k + (ln(k + )) k k= k=

224 636 Chpter 9 / Infinite Series Solution (). Solution (b). The series diverges, since ρ = lim (u k) /k 4k 5 = lim k + k + k + = > The series converges, since ρ = lim (u k) /k = lim k + k + ln(k + ) = < QUICK CHECK EXERCISES 9.5 (See pge 637 for nswers.) 4 Select between converges or diverges to fill the first blnk.. The series. Since k= k + k 8/3 b comprison with the p-series k=. (k + ) 3 /3 k+ lim = lim k + k 3 /3 k k + ( + ) 3 k 3 = 3 the series k= k3 /3 k b the test. 3. Since (k + )!/3 k+ k + lim = lim =+ k + k!/3 k k + 3 the series k= k! /3 k b the test. 4. Since ( ) /k lim = lim k + k k/ k + k = / the series k= /k k/ b the test. EXERCISE SET 9.5 Mke guess bout the convergence or divergence of the series, nd confirm our guess using the comprison test. 3. () (b) 5k k k k= k= 4 k +. () (b) k k k 4 + k k= k= 3. In ech prt, use the comprison test to show tht the series converges. 5 sin k () (b) 3 k + 5 k! k= k= 4. In ech prt, use the comprison test to show tht the series diverges. ln k k () (b) k k 3/ k= 5 Use the limit comprison test to determine whether the series converges. 4k k k 7 + k 8 9k + 6 k= k= 7. k= 5 3 k + 8. k= k= k(k + 3) (k + )(k + )(k + 5) 9. k= k= 3 8k 3k. k= (k + 3) 7 6 Use the rtio test to determine whether the series converges. If the test is inconclusive, then s so. 3 k 4 k.. 3. k! k 5k k= k= k= ( ) k k! k 4. k k 3 k + k= 7 Use the root test to determine whether the series converges. If the test is inconclusive, then s so. ( ) 3k + k ( ) k k k k= k= k 9.. ( e k ) k 5 k k= 4 True Flse Determine whether the sttement is true or flse. Eplin our nswer.. The limit comprison test decides convergence bsed on limit of the quotient of consecutive terms in series.. If lim k + (u k+ /u k ) = 5, then u k diverges. k= k=

225 9.5 The Comprison, Rtio, nd Root Tests If lim k + (k u k ) = 5, then u k converges. 4. The root test decides convergence bsed on limit of kth roots of terms in the sequence of prtil sums for series Use n method to determine whether the series converges k= 7 k k! 6. k! k 9. 3 k k= k= k= k= k= k= k= k= k= k k 3 + k= k + 7. k 5 e k 3. k= 3. k(k + ) k (k + ) 3 + k 38. k! e k k k + k k!+3 ln k 3 k 47. k= k= k= k= 36. k= k= k= k k + ( ) k k= k= 5 k k! k k cos k 3 (k + 4)! 4!k!4 k 4. k ln k k (k!) (k)! 48. k= k= k= k 5 k k k 3 + ln k e k ( k ) k k + tn k k [π(k + )] k k= k k+ 5. For wht positive vlues of α does the series k= (αk /k α ) converge? 5 5 Find the generl term of the series nd use the rtio test to show tht the series converges ! 5! 7! 53. Show tht ln < if >, nd use this result to investigte the convergence of ln k () (b) k (ln k) k= FOCUS ON CONCEPTS k= k= 54. () Mke conjecture bout the convergence of the series k= sin(π /k) b considering the locl liner pproimtion of sin t =. (b) Tr to confirm our conjecture using the limit comprison test. 55. () We will see lter tht the polnomil / is the locl qudrtic pproimtion for cos t =. Mke conjecture bout the convergence of the series [ ( )] cos k b considering this pproimtion. (b) Tr to confirm our conjecture using the limit comprison test. 56. Let k nd b k be series with positive terms. Prove: () If lim k + ( k /b k ) = nd b k converges, then k converges. (b) If lim k + ( k /b k ) =+ nd b k diverges, then k diverges. 57. Use Theorem to prove the comprison test (Theorem 9.5.). 58. Writing Wht does the rtio test tell ou bout the convergence of geometric series? Discuss similrities between geometric series nd series to which the rtio test pplies. 59. Writing Given n infinite series, discuss strteg for deciding wht convergence test to use. QUICK CHECK ANSWERS 9.5. diverges; /k /3. converges; rtio 3. diverges; rtio 4. converges; root

226 638 Chpter 9 / Infinite Series 9.6 ALTERNATING SERIES; ABSOLUTE AND CONDITIONAL CONVERGENCE Up to now we hve focused eclusivel on series with nonnegtive terms. In this section we will discuss series tht contin both positive nd negtive terms. ALTERNATING SERIES Series whose terms lternte between positive nd negtive, clled lternting series, re of specil importnce. Some emples re ( ) k+ k = k= ( ) k k = k= In generl, n lternting series hs one of the following two forms: ( ) k+ k = () k= ( ) k k = () k= where the k s re ssumed to be positive in both cses. The following theorem is the ke result on convergence of lternting series theorem (Alternting Series Test) An lternting series of either form () or form () converges if the following two conditions re stisfied: () (b) 3 k lim k = k s s 4 s 5 s 3 s = Figure 9.6. It is not essentil for condition () in Theorem 9.6. to hold for ll terms; n lternting series will converge if condition (b) is true nd condition () holds eventull. proof We will consider onl lternting series of form (). The ide of the proof is to show tht if conditions () nd (b) hold, then the sequences of even-numbered nd oddnumbered prtil sums converge to common limit S. It will then follow from Theorem 9..4 tht the entire sequence of prtil sums converges to S. Figure 9.6. shows how successive prtil sums stisfing conditions () nd (b) pper when plotted on horizontl is. The even-numbered prtil sums s,s 4,s 6,s 8,...,s n,... form n incresing sequence bounded bove b, nd the odd-numbered prtil sums s,s 3,s 5,...,s n,... form decresing sequence bounded below b. Thus, b Theorems 9..3 nd 9..4, the even-numbered prtil sums converge to some limit S E nd the odd-numbered prtil sums converge to some limit S O. To complete the proof we must show tht S E = S O. But the

227 9.6 Alternting Series; Absolute nd Conditionl Convergence 639 If n lternting series violtes condition (b) of the lternting series test, then the series must diverge b the divergence test (Theorem 9.4.). (n)-th term in the series is n, so tht s n s n = n, which cn be written s s n = s n + n However, n + nd n + s n +, so tht S O = which completes the proof. lim s n = lim (s n + n ) = S E + = S E n + n + Emple Use the lternting series test to show tht the following series converge. () ( ) k+ (b) ( ) k+ k + 3 k k(k + ) k= k= The series in prt () of Emple is clled the lternting hrmonic series. Note tht this series converges, wheres the hrmonic series diverges. Solution (). Solution (b). so k+ k = The two conditions in the lternting series test re stisfied since k = k > k + = k+ nd lim k = lim k + k + k = The two conditions in the lternting series test re stisfied since k + 4 k(k + ) = k + 4k (k + )(k + ) k + 3 k + 5k + 6 = k + 4k (k + 4k) + (k + 6) < k > k+ nd lim k + 3 k = lim k + k + k(k + ) = lim k + 3 k k + + k = APPROXIMATING SUMS OF ALTERNATING SERIES The following theorem is concerned with the error tht results when the sum of n lternting series is pproimted b prtil sum theorem If n lternting series stisfies the hpotheses of the lternting series test, nd if S is the sum of the series, then: () S lies between n two successive prtil sums; tht is, either s n S s n+ or s n+ S s n (3) (b) depending on which prtil sum is lrger. If S is pproimted b s n, then the bsolute error S s n stisfies S s n n+ (4) Moreover, the sign of the error S s n is the sme s tht of the coefficient of n+.

228 64 Chpter 9 / Infinite Series s s 4 S s 5 s 3 s Figure 9.6. proof We will prove the theorem for series of form (). Referring to Figure 9.6. nd keeping in mind our observtion in the proof of Theorem 9.6. tht the odd-numbered prtil sums form decresing sequence converging to S nd the even-numbered prtil sums form n incresing sequence converging to S, we see tht successive prtil sums oscillte from one side of S to the other in smller nd smller steps with the odd-numbered prtil sums being lrger thn S nd the even-numbered prtil sums being smller thn S. Thus, depending on whether n is even or odd, we hve s n S s n+ or s n+ S s n which proves (3). Moreover, in either cse we hve S s n s n+ s n (5) But s n+ s n =± n+ (the sign depending on whether n is even or odd). Thus, it follows from (5) tht S s n n+, which proves (4). Finll, since the odd-numbered prtil sums re lrger thn S nd the even-numbered prtil sums re smller thn S, it follows tht S s n hs the sme sign s the coefficient of n+ (verif). REMARK In words, inequlit (4) sttes tht for series stisfing the hpotheses of the lternting series test, the mgnitude of the error tht results from pproimting S b s n is t most tht of the first term tht is not included in the prtil sum. Also, note tht if > > > k >, then inequlit (4) cn be strengthened to S s n < n+..8 ln {s n } 5 5 ( ) k+ k Grph of the sequences of terms nd nth prtil sums for the lternting hrmonic series Figure n Emple series is This is illustrted in Figure Lter in this chpter we will show tht the sum of the lternting hrmonic ln = ( )k+ k + () Accepting this to be so, find n upper bound on the mgnitude of the error tht results if ln is pproimted b the sum of the first eight terms in the series. (b) Find prtil sum tht pproimtes ln to one deciml-plce ccurc (the nerest tenth). Solution (). It follows from the strengthened form of (4) tht As check, let us compute s 8 ectl. We obtin ln s 8 < 9 = <. (6) 9 s 8 = = Thus, with the help of clcultor ln s 8 = 533 ln This shows tht the error is well under the estimte provided b upper bound (6). Solution (b). For one deciml-plce ccurc, we must choose vlue of n for which ln s n.5. However, it follows from the strengthened form of (4) tht ln s n < n+ so it suffices to choose n so tht n+.5.

229 9.6 Alternting Series; Absolute nd Conditionl Convergence 64 As Emple illustrtes, the lternting hrmonic series does not provide n efficient w to pproimte ln, since too mn terms nd hence too much computtion is required to chieve resonble ccurc. Lter, we will develop better ws to pproimte logrithms. One w to find n is to use clculting utilit to obtin numericl vlues for,, 3,...until ou encounter the first vlue tht is less thn or equl to.5. If ou do this, ou will find tht it is =.5; this tells us tht prtil sum s 9 will provide the desired ccurc. Another w to find n is to solve the inequlit n +.5 lgebricll. We cn do this b tking reciprocls, reversing the sense of the inequlit, nd then simplifing to obtin n 9. Thus, s 9 will provide the required ccurc, which is consistent with the previous result. With the help of clculting utilit, the vlue of s 9 is pproimtel s 9.7 nd the vlue of ln obtined directl is pproimtel ln.69, which grees with s 9 when rounded to one deciml plce. ABSOLUTE CONVERGENCE The series does not fit in n of the ctegories studied so fr it hs mied signs but is not lternting. We will now develop some convergence tests tht cn be pplied to such series definition A series u k = u + u + +u k + k= is sid to converge bsolutel if the series of bsolute vlues u k = u + u + + u k + k= converges nd is sid to diverge bsolutel if the series of bsolute vlues diverges. Emple 3 Determine whether the following series converge bsolutel. () (b) Solution (). The series of bsolute vlues is the convergent geometric series so the given series converges bsolutel Solution (b). The series of bsolute vlues is the divergent hrmonic series so the given series diverges bsolutel

230 64 Chpter 9 / Infinite Series It is importnt to distinguish between the notions of convergence nd bsolute convergence. For emple, the series in prt (b) of Emple 3 converges, since it is the lternting hrmonic series, et we demonstrted tht it does not converge bsolutel. However, the following theorem shows tht if series converges bsolutel, then it converges theorem If the series Theorem provides w of inferring convergence of series with positive nd negtive terms from relted series with nonnegtive terms (the series of bsolute vlues). This is importnt becuse most of the convergence tests tht we hve developed ppl onl to series with nonnegtive terms. u k = u + u + + u k + k= converges, then so does the series u k = u + u + +u k + k= proof We will write the series u k s {s n } {u k } {s n } {u k } () (b) Grphs of the sequences of terms nd nth prtil sums for the series in Emple 4 Figure n n u k = k= [(u k + u k ) u k ] (7) k= We re ssuming tht u k converges, so tht if we cn show tht (u k + u k ) converges, then it will follow from (7) nd Theorem 9.4.3() tht u k converges. However, the vlue of u k + u k is either or u k, depending on the sign of u k. Thus, in ll cses it is true tht u k + u k u k But u k converges, since it is constnt times the convergent series u k ; hence (uk + u k ) converges b the comprison test. Emple 4 Show tht the following series converge. () (b) k= cos k k Solution (). Observe tht this is not n lternting series becuse the signs lternte in pirs fter the first term. Thus, we hve no convergence test tht cn be pplied directl. However, we showed in Emple 3() tht the series converges bsolutel, so Theorem implies tht it converges (Figure 9.6.4). Solution (b). With the help of clculting utilit, ou will be ble to verif tht the signs of the terms in this series vr irregulrl. Thus, we will test for bsolute convergence. The series of bsolute vlues is cos k k However, k= cos k k k

231 9.6 Alternting Series; Absolute nd Conditionl Convergence 643 But /k is convergent p-series (p = ), so the series of bsolute vlues converges b the comprison test. Thus, the given series converges bsolutel nd hence converges (Figure 9.6.4b). CONDITIONAL CONVERGENCE Although Theorem is useful tool for series tht converge bsolutel, it provides no informtion bout the convergence or divergence of series tht diverges bsolutel. For emple, consider the two series ( )k+ + (8) k 3 4 (9) k Both of these series diverge bsolutel, since in ech cse the series of bsolute vlues is the divergent hrmonic series k + However, series (8) converges, since it is the lternting hrmonic series, nd series (9) diverges, since it is constnt times the divergent hrmonic series. As mtter of terminolog, series tht converges but diverges bsolutel is sid to converge conditionll (or to be conditionll convergent). Thus, (8) is conditionll convergent series. Emple 5 In Emple (b) we used the lternting series test to show tht the series ( ) k+ k + 3 k(k + ) k= converges. Determine whether this series converges bsolutel or converges conditionll. Solution. vlues: We test the series for bsolute convergence b emining the series of bsolute k + 3 ( )k+ k(k + ) = k + 3 k(k + ) k= Principle suggests tht the series of bsolute vlues should behve like the divergent p-series with p =. To prove tht the series of bsolute vlues diverges, we will ppl the limit comprison test with k = k + 3 nd b k = k(k + ) k We obtin k k(k + 3) ρ = lim = lim k + b k k + k(k + ) = lim k + 3 k + k + = Since ρ is finite nd positive, it follows from the limit comprison test tht the series of bsolute vlues diverges. Thus, the originl series converges nd lso diverges bsolutel, nd so converges conditionll. k= THE RATIO TEST FOR ABSOLUTE CONVERGENCE Although one cnnot generll infer convergence or divergence of series from bsolute divergence, the following vrition of the rtio test provides w of deducing divergence from bsolute divergence in certin situtions. We omit the proof.

232 644 Chpter 9 / Infinite Series theorem (Rtio Test for Absolute Convergence) Let u k be series with nonzero terms nd suppose tht u k+ ρ = lim k + u k () (b) (c) If ρ<, then the series u k converges bsolutel nd therefore converges. If ρ> or if ρ =+, then the series u k diverges. If ρ =, no conclusion bout convergence or bsolute convergence cn be drwn from this test. Emple 6 converges. Use the rtio test for bsolute convergence to determine whether the series () ( ) k k k (k )! (b) ( ) k! 3 k k= k= Solution (). Thus, Tking the bsolute vlue of the generl term u k we obtin k u k = ( )k k! = k k! u k+ ρ = lim = lim k + u k k + k+ (k + )! k! k = lim k + k + = < which implies tht the series converges bsolutel nd therefore converges. Solution (b). Thus, Tking the bsolute vlue of the generl term u k we obtin u k = (k )! ( )k (k )! 3 k = 3 k u k+ [(k + ) ]! 3 k ρ = lim = lim k + u k k + 3 k+ (k )! (k + )! = lim k + 3 (k )! = 3 lim (k)(k + ) =+ k + which implies tht the series diverges. SUMMARY OF CONVERGENCE TESTS We conclude this section with summr of convergence tests tht cn be used for reference. The skill of selecting good test is developed through lots of prctice. In some instnces test m be inconclusive, so nother test must be tried.

233 Summr of Convergence Tests nme sttement comments Divergence Test (9.4.) If lim u k, then k + u k diverges. If lim u k =, then k + m not converge. u k m or Integrl Test (9.4.4) Comprison Test (9.5.) Let u k be series with positive terms. If f is function tht is decresing nd continuous on n intervl [, + ) nd such tht u k = f(k) for ll k, then + u k nd f() d k= both converge or both diverge. k= k= Let k nd b k be series with nonnegtive terms such tht b, b,..., k b k,... If b k converges, then k converges, nd if k diverges, then b k diverges. This test onl pplies to series tht hve positive terms. Tr this test when f() is es to integrte. This test onl pplies to series with nonnegtive terms. Tr this test s lst resort; other tests re often esier to ppl. Limit Comprison Test (9.5.4) Let k nd b k be series with positive terms nd let r = lim k k + b k If < r < +, then both series converge or both diverge. This is esier to ppl thn the comprison test, but still requires some skill in choosing the series b k for comprison. Let u k be series with positive terms nd suppose tht Rtio Test (9.5.5) r = u lim k+ k + u k () Series converges if r <. (b) Series diverges if r > or r = +. (c) The test is inconclusive if r =. Tr this test when u k involves fctorils or kth powers. Let u k be series with positive terms nd suppose tht Root Test (9.5.6) Alternting Series Test (9.6.) Rtio Test for Absolute Convergence (9.6.5) k r = lim k + u k () The series converges if r <. (b) The series diverges if r > or r = +. (c) The test is inconclusive if r =. If k > for k =,, 3,..., then the series converge if the following conditions hold: () 3... (b) lim k = k + Let u k be series with nonzero terms nd suppose tht u r = lim k+ k + u k () The series converges bsolutel if r <. (b) The series diverges if r > or r = +. (c) The test is inconclusive if r =. Tr this test when u k involves kth powers. This test pplies onl to lternting series. The series need not hve positive terms nd need not be lternting to use this test.

234 646 Chpter 9 / Infinite Series QUICK CHECK EXERCISES 9.6 (See pge 648 for nswers.). Wht chrcterizes n lternting series?. () The series ( ) k+ k= converges b the lternting series test since nd. (b) If S = k ( ) k+ nd s k 9 = k= then S s 9 <. 9 ( ) k+ 3. Clssif ech sequence s conditionll convergent, bsolutel convergent, or divergent. () ( ) k+ k : k= k= k (b) (c) (d) ( ) k 3k 9k + 5 : ( ) k k(k + ) : ( ) k+ 4 : k 3 k= k= k= 4. Given tht (k + ) 4 /4 k+ lim = lim k + k 4 /4 k k + ( + ) 4 k 4 = 4 is the series k= ( )k k 4 /4 k conditionll convergent, bsolutel convergent, or divergent? EXERCISE SET 9.6 C CAS Show tht the series converges b confirming tht it stisfies the hpotheses of the lternting series test (Theorem 9.6.). ( ) k+.. ( ) k+ k k + 3 k k= 3 6 Determine whether the lternting series converges; justif our nswer. 3. ( ) k+ k + 4. ( ) k+ k + 3k + k= k= k + 5. ( ) k+ e k 6. ( ) k ln k k k= 7 Use the rtio test for bsolute convergence (Theorem 9.6.5) to determine whether the series converges or diverges. If the test is inconclusive, s so. ( 7. 3 ) k k+ k 8. ( ) 5 k! k= k= k+ 3k 9. ( ). ( ) k k k 5 k k= k=. ( ) k k3 k+ kk. ( ) e k k! k= 3 8 Clssif ech series s bsolutel convergent, conditionll convergent, or divergent. ( ) k+ ( ) k+ ( 4) k k k 4/3 k k= k= k= k=3 k= k= ( ) k+ k! k= k= 7. ( ) k+ k + k(k + 3) sin kπ k= k= k= cos kπ k.. 8. k=3 ( ) k+ k k 3 + k= k= ( ) k 4. k ln k k= k= ( ) k 6. ln k k= ( ) k+ k! (k )! 8. k= sin k k 3 ( ) k k(k + ) k cos kπ k + ( ) k= k+ 3k ( ) k ln k k k True Flse Determine whether the sttement is true or flse. Eplin our nswer. 9. An lternting series is one whose terms lternte between even nd odd. 3. If series stisfies the hpothesis of the lternting series test, then the sequence of prtil sums of the series oscilltes between overestimtes nd underestimtes for the sum of the series. 3. If series converges, then either it converges bsolutel or it converges conditionll. 3. If (u k ) converges, then u k converges bsolutel.

235 9.6 Alternting Series; Absolute nd Conditionl Convergence Ech series stisfies the hpotheses of the lternting series test. For the stted vlue of n, find n upper bound on the bsolute error tht results if the sum of the series is pproimted b the nth prtil sum. ( ) k+ ( ) k+ 33. ; n = ; n = 5 k k! k= k= k= ( ) k+ k ; n = 99 ( ) k+ (k + ) ln(k + ) ; n = Ech series stisfies the hpotheses of the lternting series test. Find vlue of n for which the nth prtil sum is ensured to pproimte the sum of the series to the stted ccurc. ( ) k+ 37. ; error <. k k= ( ) k+ 38. ; error <. k! k= k= k= k= ( ) k+ ; two deciml plces k ( ) k+ ; one deciml plce (k + ) ln(k + ) 4 4 Find n upper bound on the bsolute error tht results if s is used to pproimte the sum of the given geometric series. Compute s rounded to four deciml plces nd compre this vlue with the ect sum of the series Ech series stisfies the hpotheses of the lternting series test. Approimte the sum of the series to two deciml-plce ccurc. C 43. 3! + 5! 7! + 44.! + 4! 6! FOCUS ON CONCEPTS 47. The purpose of this eercise is to show tht the error bound in prt (b) of Theorem 9.6. cn be overl conservtive in certin cses. () Use CAS to confirm tht π 4 = (b) Use the CAS to show tht (π/4) s 5 <. (c) According to the error bound in prt (b) of Theorem 9.6., wht vlue of n is required to ensure tht (π/4) s n <? 48. Prove: If series k converges bsolutel, then the series k converges. 49. () Find emples to show tht if k converges, then k m diverge or converge. (b) Find emples to show tht if k converges, then k m diverge or converge. 5. Let u k be series nd define series p k nd q k so tht { { uk, u p k = k >, uk nd q, u k k = u k, u k < () Show tht u k converges bsolutel if nd onl if pk nd q k both converge. (b) Show tht if one of p k or q k converges nd the other diverges, then u k diverges. (c) Show tht if u k converges conditionll, then both pk nd q k diverge. 5. It cn be proved tht the terms of n conditionll convergent series cn be rerrnged to give either divergent series or conditionll convergent series whose sum is n given number S. For emple, we stted in Emple tht ln = Show tht we cn rerrnge this series so tht its sum is ln b rewriting it s ( ) ( ) ( ) + [Hint: Add the first two terms in ech grouping.] 5 54 Eercise 5 illustrtes tht one of the nunces of conditionl convergence is tht the sum of series tht converges conditionll depends on the order tht the terms of the series re summed. Absolutel convergent series re more dependble, however. It cn be proved tht n series tht is constructed from n bsolutel convergent series b rerrnging the terms will lso be bsolutel convergent nd hs the sme sum s the originl series. Use this fct together with prts () nd (b) of Theorem in these eercises. 5. It ws stted in Eercise 35 of Section 9.4 tht π 6 = Use this to show tht π 8 = Use the series for π /6 given in the preceding eercise to show tht π =

236 648 Chpter 9 / Infinite Series 54. It ws stted in Eercise 35 of Section 9.4 tht π 4 9 = Use this to show tht π 4 96 = Writing Consider the series Determine whether this series converges nd use this series s n emple in discussion of the importnce of hpotheses () nd (b) of the lternting series test (Theorem 9.6.). 56. Writing Discuss the ws tht conditionl convergence is conditionl. In prticulr, describe how one could rerrnge the terms of conditionll convergent series u k so tht the resulting series diverges, either to + or to. [Hint: See Eercise 5.] QUICK CHECK ANSWERS 9.6. Terms lternte between positive nd negtive.. () 4 9 k ; lim (k + ) k + k = (b) 3. () conditionll convergent (b) divergent (c) bsolutel convergent (d) conditionll convergent 4. bsolutel convergent 9.7 MACLAURIN AND TAYLOR POLYNOMIALS In locl liner pproimtion the tngent line to the grph of function is used to obtin liner pproimtion of the function ner the point of tngenc. In this section we will consider how one might improve on the ccurc of locl liner pproimtions b using higher-order polnomils s pproimting functions. We will lso investigte the error ssocited with such pproimtions. LOCAL QUADRATIC APPROXIMATIONS Recll from Formul () in Section 3.5 tht the locl liner pproimtion of function f t is f() f( ) + f ( )( ) () In this formul, the pproimting function p() = f( ) + f ( )( ) Figure 9.7. f Locl liner pproimtion is first-degree polnomil stisfing p( ) = f( ) nd p ( ) = f ( ) (verif). Thus, the locl liner pproimtion of f t hs the propert tht its vlue nd the vlue of its first derivtive mtch those of f t. If the grph of function f hs pronounced bend t, then we cn epect tht the ccurc of the locl liner pproimtion of f t will decrese rpidl s we progress w from (Figure 9.7.). One w to del with this problem is to pproimte the function f t b polnomil p of degree with the propert tht the vlue of p nd the vlues of its first two derivtives mtch those of f t. This ensures tht the grphs of f nd p not onl hve the sme tngent line t, but the lso bend in the sme direction t (both concve up or concve down). As result, we cn epect tht the grph of p will remin close to the grph of f over lrger intervl round thn the grph of the locl liner pproimtion. The polnomil p is clled the locl qudrtic pproimtion of f t =. To illustrte this ide, let us tr to find formul for the locl qudrtic pproimtion of function f t =. This pproimtion hs the form f() c + c + c () where c, c, nd c must be chosen so tht the vlues of p() = c + c + c

237 nd its first two derivtives mtch those of f t. Thus, we wnt 9.7 Mclurin nd Tlor Polnomils 649 p() = f(), p () = f (), p () = f () (3) But the vlues of p(), p (), nd p () re s follows: Thus, it follows from (3) tht p() = c + c + c p() = c p () = c + c p () = c p () = c p () = c c = f(), c = f (), c = f () nd substituting these in () ields the following formul for the locl qudrtic pproimtion of f t = : f() f() + f () + f () (4) Emple Find the locl liner nd qudrtic pproimtions of e t =, nd grph e nd the two pproimtions together. = + + = e Figure 9.7. = + Solution. If we let f() = e, then f () = f () = e ; nd hence f() = f () = f () = e = Thus, from (4) the locl qudrtic pproimtion of e t = is e + + nd the locl liner pproimtion (which is the liner prt of the locl qudrtic pproimtion) is e + The grphs of e nd the two pproimtions re shown in Figure As epected, the locl qudrtic pproimtion is more ccurte thn the locl liner pproimtion ner =. MACLAURIN POLYNOMIALS It is nturl to sk whether one cn improve on the ccurc of locl qudrtic pproimtion b using polnomil of degree 3. Specificll, one might look for polnomil of degree 3 with the propert tht its vlue nd the vlues of its first three derivtives mtch Colin Mclurin ( ) Scottish mthemticin. Mclurin s fther, minister, died when the bo ws onl si months old, nd his mother when he ws nine ers old. He ws then rised b n uncle who ws lso minister. Mclurin entered Glsgow Universit s divinit student but switched to mthemtics fter one er. He received his Mster s degree t ge 7 nd, in spite of his outh, begn teching t Mrischl College in Aberdeen, Scotlnd. He met Isc Newton during visit to London in 79 nd from tht time on becme Newton s disciple. During tht er, some of Newton s nltic methods were bitterl ttcked b mjor mthemticins nd much of Mclurin s importnt mthemticl work resulted from his efforts to defend Newton s ides geometricll. Mclurin s work, A Tretise of Fluions (74), ws the first sstemtic formultion of Newton s methods. The tretise ws so crefull done tht it ws stndrd of mthemticl rigor in clculus until the work of Cuch in 8. Mclurin ws lso n outstnding eperimentlist; he devised numerous ingenious mechnicl devices, mde importnt stronomicl observtions, performed cturil computtions for insurnce societies, nd helped to improve mps of the islnds round Scotlnd. [Imge: Bettmnn/Corbis Imges]

238 65 Chpter 9 / Infinite Series those of f t point; nd if this provides n improvement in ccurc, wh not go on to polnomils of even higher degree? Thus, we re led to consider the following generl problem problem Given function f tht cn be differentited n times t =, find polnomil p of degree n with the propert tht the vlue of p nd the vlues of its first n derivtives mtch those of f t. We will begin b solving this problem in the cse where =. Thus, we wnt polnomil p() = c + c + c + c c n n (5) such tht f() = p(), f () = p (), f () = p (),..., f (n) () = p (n) () (6) But p() = c + c + c + c c n n p () = c + c + 3c 3 + +nc n n p () = c + 3 c 3 + +n(n )c n n p () = 3 c 3 + +n(n )(n )c n n 3 Thus, to stisf (6) we must hve. p (n) () = n(n )(n ) ()c n f() = p() = c f () = p () = c f () = p () = c =!c f () = p () = 3 c 3 = 3!c 3. f (n) () = p (n) () = n(n )(n ) ()c n = n!c n which ields the following vlues for the coefficients of p(): c = f(), c = f (), c = f ()!, c 3 = f (),..., c n = f (n) () 3! n! The polnomil tht results b using these coefficients in (5) is clled the nth Mclurin polnomil for f. Locl liner pproimtions nd locl qudrtic pproimtions t = of function f re specil cses of the McLurin polnomils for f. Verif tht f() p () is the locl liner pproimtion of f t =, nd f() p () is the locl qudrtic pproimtion t = definition If f cn be differentited n times t, then we define the nth Mclurin polnomil for f to be p n () = f() + f () + f ()! + f () 3! f (n) () n (7) n! Note tht the polnomil in (7) hs the propert tht its vlue nd the vlues of its first n derivtives mtch the vlues of f nd its first n derivtives t =.

239 9.7 Mclurin nd Tlor Polnomils 65 Emple Find the Mclurin polnomils p, p, p, p 3, nd p n for e. Solution. Let f() = e. Thus, f () = f () = f () = =f (n) () = e nd f() = f () = f () = f () = =f (n) () = e = Therefore, p () = f() = p () = f() + f () = + p () = f() + f () + f () = + +!! = + + p 3 () = f() + f () + f ()! + f () 3 3! 5 4 = e p 3 () p () = + +! + 3 3! = p n () = f() + f () + f ()! = + + n + +! n! + + f (n) () n n! 3 p () p () Figure Figure shows the grph of e (in blue) nd the grph of the first four Mclurin polnomils. Note tht the grphs of p (), p (), nd p 3 () re virtull indistinguishble from the grph of e ner =, so these polnomils re good pproimtions of e for ner. However, the frther is from, the poorer these pproimtions become. This is tpicl of the Mclurin polnomils for function f(); the provide good pproimtions of f() ner, but the ccurc diminishes s progresses w from. It is usull the cse tht the higher the degree of the polnomil, the lrger the intervl on which it provides specified ccurc. Accurc issues will be investigted lter. Augustin Louis Cuch ( ) French mthemticin. Cuch s erl eduction ws cquired from his fther, brrister nd mster of the clssics. Cuch entered L Ecole Poltechnique in 85 to stud engineering, but becuse of poor helth, ws dvised to concentrte on mthemtics. His mjor mthemticl work begn in 8 with series of brillint solutions to some difficult outstnding problems. In 84 he wrote tretise on integrls tht ws to become the bsis for modern comple vrible theor; in 86 there followed clssic pper on wve propgtion in liquids tht won prize from the French Acdem; nd in 8 he wrote pper tht formed the bsis of modern elsticit theor. Cuch s mthemticl contributions for the net 35 ers were brillint nd stggering in quntit, over 7 ppers filling 6 modern volumes. Cuch s work initited the er of modern nlsis. He brought to mthemtics stndrds of precision nd rigor undremed of b Leibniz nd Newton. Cuch s life ws inetricbl tied to the politicl uphevls of the time. A strong prtisn of the Bourbons, he left his wife nd children in 83 to follow the Bourbon king Chrles X into eile. For his lolt he ws mde bron b the e-king. Cuch eventull returned to Frnce, but refused to ccept universit position until the government wived its requirement tht he tke lolt oth. It is difficult to get cler picture of the mn. Devoutl Ctholic, he sponsored chritble work for unwed mothers, criminls, nd relief for Irelnd. Yet other spects of his life cst him in n unfvorble light. The Norwegin mthemticin Abel described him s, md, infinitel Ctholic, nd bigoted. Some writers prise his teching, et others s he rmbled incoherentl nd, ccording to report of the d, he once devoted n entire lecture to etrcting the squre root of seventeen to ten deciml plces b method well known to his students. In n event, Cuch is undenibl one of the gretest minds in the histor of science. [Imge:

240 65 Chpter 9 / Infinite Series Emple 3 Find the nth Mclurin polnomils for () sin (b) cos Solution (). In the Mclurin polnomils for sin, onl the odd powers of pper eplicitl. To see this, let f() = sin ; thus, f() = sin f() = f () = cos f () = f () = sin f () = f () = cos f () = Since f (4) () = sin = f(), the pttern,,, will repet s we evlute successive derivtives t. Therefore, the successive Mclurin polnomils for sin re p () = p () = + p () = + + p 3 () = ! p 4 () = ! + p 5 () = ! ! C Figure p () p 5 () c = sin p 3 () p 7 () p 6 () = ! ! + p 7 () = ! ! + 7 7! Becuse of the zero terms, ech even-order Mclurin polnomil [fter p ()] is the sme s the preceding odd-order Mclurin polnomil. Tht is, p k+ () = p k+ () = 3 3! + 5 5! 7 k+ 7! + +( )k (k =,,,...) (k + )! The grphs of sin,p (), p 3 (), p 5 (), nd p 7 () re shown in Figure Solution (b). In the Mclurin polnomils for cos, onl the even powers of pper eplicitl; the computtions re similr to those in prt (). The reder should be ble to show tht p () = p () = p () = p 3 () =! C Figure p 4 () p () c p () = cos p 6 () p 4 () = p 5 () =! + 4 4! p 6 () = p 7 () =! + 4 4! 6 6! In generl, the Mclurin polnomils for cos re given b p k () = p k+ () =! + 4 4! 6 k + +( )k (k =,,,...) 6! (k)! The grphs of cos, p (), p (), p 4 (), nd p 6 () re shown in Figure

241 9.7 Mclurin nd Tlor Polnomils 653 TAYLOR POLYNOMIALS Up to now we hve focused on pproimting function f in the vicinit of =. Now we will consider the more generl cse of pproimting f in the vicinit of n rbitrr domin vlue. The bsic ide is the sme s before; we wnt to find n nth-degree polnomil p with the propert tht its vlue nd the vlues of its first n derivtives mtch those of f t. However, rther thn epressing p() in powers of, it will simplif the computtions if we epress it in powers of ; tht is, p() = c + c ( ) + c ( ) + +c n ( ) n (8) We will leve it s n eercise for ou to imitte the computtions used in the cse where = to show tht c = f( ), c = f ( ), c = f ( )!, c 3 = f ( ),..., c n = f (n) ( ) 3! n! Substituting these vlues in (8) we obtin polnomil clled the nth Tlor polnomil bout = for f. Locl liner pproimtions nd locl qudrtic pproimtions t = of function f re specil cses of the Tlor polnomils for f. Verif tht f() p () is the locl liner pproimtion of f t =, nd f() p () is the locl qudrtic pproimtion t = definition If f cn be differentited n times t, then we define the nth Tlor polnomil for f bout = to be p n () = f( ) + f ( )( ) + f ( ) ( )! + f ( ) 3! ( ) f (n) ( ) ( ) n (9) n! The Mclurin polnomils re the specil cses of the Tlor polnomils in which =. Thus, theorems bout Tlor polnomils lso ppl to Mclurin polnomils. Emple 4 Find the first four Tlor polnomils for ln bout =. Solution. Let f() = ln. Thus, f() = ln f() = ln f () = / f () = / f () = / f () = /4 f () = / 3 f () = /4 Brook Tlor (685 73) English mthemticin. Tlor ws born of well-to-do prents. Musicins nd rtists were entertined frequentl in the Tlor home, which undoubtedl hd lsting influence on him. In lter ers, Tlor published definitive work on the mthemticl theor of perspective nd obtined mjor mthemticl results bout the vibrtions of strings. There lso eists n unpublished work, On Musick, tht ws intended to be prt of joint pper with Isc Newton. Tlor s life ws scrred with unhppiness, illness, nd trged. Becuse his first wife ws not rich enough to suit his fther, the two men rgued bitterl nd prted ws. Subsequentl, his wife died in childbirth. Then, fter he remrried, his second wife lso died in childbirth, though his dughter survived. Tlor s most productive period ws from 74 to 79, during which time he wrote on wide rnge of subjects mgnetism, cpillr ction, thermometers, perspective, nd clculus. In his finl ers, Tlor devoted his writing efforts to religion nd philosoph. According to Tlor, the results tht ber his nme were motivted b coffeehouse converstions bout works of Newton on plnetr motion nd works of Hlle ( Hlle s comet ) on roots of polnomils. Unfortuntel, Tlor s writing stle ws so terse nd hrd to understnd tht he never received credit for mn of his innovtions. [Imge:

242 654 Chpter 9 / Infinite Series = ln p 3 () p () p () p () Substituting in (9) with = ields p () = f() = ln p () = f() + f ()( ) = ln + ( ) p () = f() + f ()( ) + f () ( ) = ln +! ( ) ( ) 8 p 3 () = f() + f ()( ) + f ()! = ln + ( ) 8 ( ) + ( )3 4 ( ) + f () ( ) 3 3! The grph of ln (in blue) nd its first four Tlor polnomils bout = re shown in Figure As epected, these polnomils produce their best pproimtions of ln Figure ner. SIGMA NOTATION FOR TAYLOR AND MACLAURIN POLYNOMIALS Frequentl, we will wnt to epress Formul (9) in sigm nottion. To do this, we use the nottion f (k) ( ) to denote the kth derivtive of f t =, nd we mke the convention tht f () ( ) denotes f( ). This enbles us to write n f (k) ( ) ( ) k = f( ) + f ( )( ) k! k= k= + f ( )! ( ) + + f (n) ( ) ( ) n () n! In prticulr, we cn write the nth Mclurin polnomil for f() s n f (k) () k = f() + f () + f () + + f (n) () n () k!! n! Emple 5 Find the nth Mclurin polnomil for nd epress it in sigm nottion. TECHNOLOGY MASTERY Computer lgebr sstems hve commnds for generting Tlor polnomils of n specified degree. If ou hve CAS, use it to find some of the Mclurin nd Tlor polnomils in Emples 3, 4, nd 5. Solution. s follows: Let f() = /( ). The vlues of f nd its first k derivtives t = re f() = f() = =! f () = ( ) f () = =! f () = ( ) 3 f () = =! f () = 3 ( ) 4 f () = 3! f (4) () = 4 3 ( ) 5 f (4) () = 4!.. f (k) k! () = f (k) () = k! ( ) k+

243 9.7 Mclurin nd Tlor Polnomils 655 Thus, substituting f (k) () = k! into Formul () ields the nth Mclurin polnomil for /( ): n p n () = k = n (n =,,,...) k= Emple 6 sigm nottion. Find the nth Tlor polnomil for / bout = nd epress it in Solution. Let f() = /. The computtions re similr to those in Emple 5. We leve it for ou to show tht f() =, f () =, f () =!, f () = 3!, f (4) () = 4!,..., f (k) () = ( ) k k! Thus, substituting f (k) () = ( ) k k! into Formul () with = ields the nth Tlor polnomil for /: n ( ) k ( ) k = ( ) + ( ) ( ) 3 + +( ) n ( ) n k= THE nth REMAINDER It will be convenient to hve nottion for the error in the pproimtion f() p n (). Accordingl, we will let R n () denote the difference between f() nd its nth Tlor polnomil; tht is, This cn lso be written s R n () = f() p n () = f() f() = p n () + R n () = n k= n k= f (k) ( ) ( ) k () k! f (k) ( ) ( ) k + R n () (3) k! The function R n () is clled the nth reminder for the Tlor series of f, nd Formul (3) is clled Tlor s formul with reminder. Finding bound for R n () gives n indiction of the ccurc of the pproimtion p n () f(). The following theorem, which is proved in Appendi D, provides such bound. The bound for R n () in (4) is clled the Lgrnge error bound theorem (The Reminder Estimtion Theorem) If the function f cn be differentited n + times on n intervl contining the number, nd if M is n upper bound for f (n+) () on the intervl, tht is, f (n+) () M for ll in the intervl, then R n () M (n + )! n+ (4) for ll in the intervl.

244 656 Chpter 9 / Infinite Series Use n nth Mclurin polnomil for e to pproimte e to five deciml- Emple 7 plce ccurc. Solution. We note first tht the eponentil function e hs derivtives of ll orders for ever rel number. From Emple, the nth Mclurin polnomil for e is n k k! = + + n + +! n! from which we hve k= e = e n k= k k! = + +! + + n! Thus, our problem is to determine how mn terms to include in Mclurin polnomil for e to chieve five deciml-plce ccurc; tht is, we wnt to choose n so tht the bsolute vlue of the nth reminder t = stisfies R n ().5 To determine n we use the Reminder Estimtion Theorem with f() = e, =, =, nd the intervl [, ]. In this cse it follows from (4) tht R n () M (n + )! n+ = M (n + )! where M is n upper bound on the vlue of f (n+) () = e for in the intervl [, ]. However, e is n incresing function, so its mimum vlue on the intervl [, ] occurs t = ; tht is, e e on this intervl. Thus, we cn tke M = e in (5) to obtin e R n () (6) (n + )! Unfortuntel, this inequlit is not ver useful becuse it involves e, which is the ver quntit we re tring to pproimte. However, if we ccept tht e<3, then we cn replce (6) with the following less precise, but more esil pplied, inequlit: R n () 3 (n + )! Thus, we cn chieve five deciml-plce ccurc b choosing n so tht 3.5 (n + )! or (n + )! 6, Since 9! =36,88 nd! =3,68,8, the smllest vlue of n tht meets this criterion is n = 9. Thus, to five deciml-plce ccurc e + +! + 3! + 4! + 5! + 6! + 7! + 8! + 9!.788 As check, clcultor s -digit representtion of e is e , which grees with the preceding pproimtion when rounded to five deciml plces. (5) Emple 8 Use the Reminder Estimtion Theorem to find n intervl contining = throughout which f() = cos cn be pproimted b p() = ( /!) to three deciml-plce ccurc. Solution. We note first tht f() = cos hs derivtives of ll orders for ever rel number, so the first hpothesis of the Reminder Estimtion Theorem is stisfied over n intervl tht we choose. The given polnomil p() is both the second nd the third

245 9.7 Mclurin nd Tlor Polnomils Figure = f () p() Mclurin polnomil for cos ; we will choose the degree n of the polnomil to be s lrge s possible, so we will tke n = 3. Our problem is to determine n intervl on which the bsolute vlue of the third reminder t stisfies R 3 ().5 We will use the Reminder Estimtion Theorem with f() = cos, n = 3, nd =. It follows from (4) tht R 3 () M (3 + )! 3+ = M 4 4 where M is n upper bound for f (4) () = cos. Since cos for ever rel number, we cn tke M = in (7) to obtin R 3 () 4 (8) 4 Thus we cn chieve three deciml-plce ccurc b choosing vlues of for which or.339 so the intervl [.339,.339] is one option. We cn check this nswer b grphing f() p() over the intervl [.339,.339] (Figure 9.7.7). (7) QUICK CHECK EXERCISES 9.7 (See pge 659 for nswers.). If f cn be differentited three times t, then the third Mclurin polnomil for f is p 3 () =.. The third Mclurin polnomil for f() = e is p 3 () = If f() = 3, f () = 4, nd f () =, then the second Tlor polnomil for f bout = isp () =. 4. The third Tlor polnomil for f() = 5 bout = is p 3 () = + ( + ) + ( + ) + ( + ) 3 5. () If function f hs nth Tlor polnomil p n () bout =, then the nth reminder R n () is defined b R n () =. (b) Suppose tht function f cn be differentited five times on n intervl contining = nd tht f (5) () for ll in the intervl. Then the fourth reminder stisfies R 4 () for ll in the intervl. EXERCISE SET 9.7 Grphing Utilit In ech prt, find the locl qudrtic pproimtion of f t =, nd use tht pproimtion to find the locl liner pproimtion of f t. Use grphing utilit to grph f nd the two pproimtions on the sme screen.. () f() = e ; = (b) f() = cos ; =. () f() = sin ; = π/ (b) f() = ; = 3. () Find the locl qudrtic pproimtion of t =. (b) Use the result obtined in prt () to pproimte., nd compre our pproimtion to tht produced directl b our clculting utilit. [Note: See Emple of Section 3.5.] 4. () Find the locl qudrtic pproimtion of cos t =. (b) Use the result obtined in prt () to pproimte cos, nd compre the pproimtion to tht produced directl b our clculting utilit. 5. Use n pproprite locl qudrtic pproimtion to pproimte tn 6, nd compre the result to tht produced directl b our clculting utilit. 6. Use n pproprite locl qudrtic pproimtion to pproimte 36.3, nd compre the result to tht produced directl b our clculting utilit.

246 658 Chpter 9 / Infinite Series 7 6 Find the Mclurin polnomils of orders n =,,, 3, nd 4, nd then find the nth Mclurin polnomils for the function in sigm nottion. 7. e 8. e 9. cos π. sin π. ln( + ) cosh 4. sinh 5. sin 6. e 7 4 Find the Tlor polnomils of orders n =,,, 3, nd 4 bout =, nd then find the nth Tlor polnomil for the function in sigm nottion. 7. e ; = 8. e ; = ln 9. ; =. + ; = 3. sin π; =. cos ; = π 3. ln ; = 4. ln ; = e 5. () Find the third Mclurin polnomil for f() = (b) Find the third Tlor polnomil bout = for f() = + ( ) ( ) + ( ) 3 6. () Find the nth Mclurin polnomil for f() = c + c + c + +c n n (b) Find the nth Tlor polnomil bout = for f() = c + c ( ) + c ( ) + +c n ( ) n 7 3 Find the first four distinct Tlor polnomils bout =, nd use grphing utilit to grph the given function nd the Tlor polnomils on the sme screen. 7. f() = e ; = 8. f() = sin ; = π/ 9. f() = cos ; = π 3. ln( + ); = 3 34 True Flse Determine whether the sttement is true or flse. Eplin our nswer. 3. The eqution of tngent line to differentible function is first-degree Tlor polnomil for tht function. 3. The grph of function f nd the grph of its Mclurin polnomil hve common -intercept. 33. If p 6 () is the sith-degree Tlor polnomil for function f bout =, then p (4) 6 ( ) = 4!f (4) ( ). 34. If p 4 () is the fourth-degree Mclurin polnomil for e, then e p 4 () 9 5! Use the method of Emple 7 to pproimte the given epression to the specified ccurc. Check our nswer to tht produced directl b our clculting utilit. 35. e; four deciml-plce ccurc 36. /e; three deciml-plce ccurc FOCUS ON CONCEPTS 37. Which of the functions grphed in the following figure is most likel to hve p() = + s its secondorder Mclurin polnomil? Eplin our resoning. I II III IV 38. Suppose tht the vlues of function f nd its first three derivtives t = re f() =, f () = 3, f () =, f () = 6 Find s mn Tlor polnomils for f s ou cn bout =. 39. Let p () nd p () be the locl liner nd locl qudrtic pproimtions of f() = e sin t =. () Use grphing utilit to generte the grphs of f(), p (), nd p () on the sme screen for. (b) Construct tble of vlues of f(), p (), nd p () for =.,.75,.5,.5,,.5,.5,.75,.. Round the vlues to three deciml plces. (c) Generte the grph of f() p (), nd use the grph to determine n intervl on which p () pproimtes f() with n error of t most ±.. [Suggestion: Review the discussion relting to Figure ] (d) Generte the grph of f() p (), nd use the grph to determine n intervl on which p () pproimtes f() with n error of t most ±.. 4. () The ccompning figure shows sector of rdius r nd centrl ngle α. Assuming tht the ngle α is smll, use the locl qudrtic pproimtion of cos α t α = to show tht rα /. (b) Assuming tht the Erth is sphere of rdius 4 mi, use the result in prt () to pproimte the mimum mount b which mi rc long the equtor will diverge from its chord. r r Figure E-4

247 9.8 Mclurin nd Tlor Series; Power Series () Find n intervl [,b] over which e cn be pproimted b + + ( /!) to three deciml-plce ccurc throughout the intervl. (b) Check our nswer in prt () b grphing ) e ( + +! over the intervl ou obtined. 4. Show tht the nth Tlor polnomil for sinh bout = ln 4 is n 6 ( ) k ( ln 4) k 8k! k= Use the Reminder Estimtion Theorem to find n intervl contining = over which f() cn be pproimted b p() to three deciml-plce ccurc throughout the intervl. Check our nswer b grphing f() p() over the intervl ou obtined. 43. f() = sin ; p() = 3 3! 44. f() = cos ; p() =! + 4 4! 45. f() = + ; p() = f() = ln( + ); p() = QUICK CHECK ANSWERS 9.7. f() + f () + f ()! 5. () f() p n () (b) f () 3. ; ; ; ( ) + 5( ) 4. ; 5; ; 3 3! 9.8 MACLAURIN AND TAYLOR SERIES; POWER SERIES Recll from the lst section tht the nth Tlor polnomil p n () t = for function f ws defined so its vlue nd the vlues of its first n derivtives mtch those of f t. This being the cse, it is resonble to epect tht for vlues of ner the vlues of p n () will become better nd better pproimtions of f() s n increses, nd m possibl converge to f() s n +. We will eplore this ide in this section. MACLAURIN AND TAYLOR SERIES In Section 9.7 we defined the nth Mclurin polnomil for function f s n k= f (k) () k! k = f() + f () + f ()! nd the nth Tlor polnomil for f bout = s n k= f (k) ( ) ( ) k = f( ) + f ( )( ) k! + f ( )! + + f (n) () n n! ( ) + + f (n) ( ) ( ) n n! It is not big step to etend the notions of Mclurin nd Tlor polnomils to series b not stopping the summtion inde t n. Thus, we hve the following definition.

248 66 Chpter 9 / Infinite Series 9.8. definition If f hs derivtives of ll orders t, then we cll the series k= f (k) ( ) k! ( ) k = f( ) + f ( )( ) + f ( ) ( )! the Tlor series for f bout =. becomes k= f (k) () k! k = f() + f () + f ()! in which cse we cll it the Mclurin series for f. + + f (k) ( ) ( ) k + () k! In the specil cse where =, this series + + f (k) () k + () k! Note tht the nth Mclurin nd Tlor polnomils re the nth prtil sums for the corresponding Mclurin nd Tlor series. Emple Solution (). for e is Find the Mclurin series for () e (b) sin (c) cos (d) In Emple of Section 9.7 we found tht the nth Mclurin polnomil n k p n () = k! = + + n + +! n! k= Thus, the Mclurin series for e is k k! = + + k + +! k! + k= Solution (b). In Emple 3() of Section 9.7 we found tht the Mclurin polnomils for sin re given b p k+ () = p k+ () = 3 3! + 5 5! 7 7! + +( )k k+ (k + )! (k =,,,...) Thus, the Mclurin series for sin is ( ) k k+ (k + )! = 3 3! + 5 5! 7 k+ 7! + +( )k (k + )! + k= Solution (c). In Emple 3(b) of Section 9.7 we found tht the Mclurin polnomils for cos re given b p k () = p k+ () =! + 4 4! 6 k + +( )k 6! (k)! Thus, the Mclurin series for cos is ( ) k k (k)! =! + 4 4! 6 6! k= (k =,,,...) + +( )k k (k)! +

249 9.8 Mclurin nd Tlor Series; Power Series 66 Solution (d). In Emple 5 of Section 9.7 we found tht the nth Mclurin polnomil for /( ) is n p n () = k = n (n =,,,...) k= Thus, the Mclurin series for /( ) is k = k + k= Emple Find the Tlor series for / bout =. Solution. In Emple 6 of Section 9.7 we found tht the nth Tlor polnomil for / bout = is n ( ) k ( ) k = ( ) + ( ) ( ) 3 + +( ) n ( ) n k= Thus, the Tlor series for / bout = is ( ) k ( ) k = ( ) + ( ) ( ) 3 + +( ) k ( ) k + k= POWER SERIES IN Mclurin nd Tlor series differ from the series tht we hve considered in Sections 9.3 to 9.6 in tht their terms re not merel constnts, but insted involve vrible. These re emples of power series, which we now define. If c, c, c,...re constnts nd is vrible, then series of the form c k k = c + c + c + +c k k + (3) k= is clled power series in. Some emples re k = k= k k! = + +! + 3 3! + ( ) k k (k)! =! + 4 4! 6 6! + k= k= From Emple, these re the Mclurin series for the functions /( ),e, nd cos, respectivel. Indeed, ever Mclurin series k= f (k) () k! is power series in. k = f() + f () + f ()! + + f (k) () k + k!

250 66 Chpter 9 / Infinite Series RADIUS AND INTERVAL OF CONVERGENCE If numericl vlue is substituted for in power series c k k, then the resulting series of numbers m either converge or diverge. This leds to the problem of determining the set of -vlues for which given power series converges; this is clled its convergence set. Observe tht ever power series in converges t =, since substituting this vlue in (3) produces the series c whose sum is c. In some cses = m be the onl number in the convergence set; in other cses the convergence set is some finite or infinite intervl contining =. This is the content of the following theorem, whose proof will be omitted theorem For n power series in, ectl one of the following is true: () The series converges onl for =. (b) The series converges bsolutel (nd hence converges) for ll rel vlues of. (c) The series converges bsolutel (nd hence converges) for ll in some finite open intervl ( R,R) nd diverges if < R or >R. At either of the vlues = R or = R, the series m converge bsolutel, converge conditionll, or diverge, depending on the prticulr series. This theorem sttes tht the convergence set for power series in is lws n intervl centered t = (possibl just the vlue = itself or possibl infinite). For this reson, the convergence set of power series in is clled the intervl of convergence. In the cse where the convergence set is the single vlue = we s tht the series hs rdius of convergence, in the cse where the convergence set is (, + ) we s tht the series hs rdius of convergence +, nd in the cse where the convergence set etends between R nd R we s tht the series hs rdius of convergence R (Figure 9.8.). Diverges Diverges Rdius of convergence R = Converges Rdius of convergence R = + Figure 9.8. Diverges R Converges R Diverges Rdius of convergence R FINDING THE INTERVAL OF CONVERGENCE The usul procedure for finding the intervl of convergence of power series is to ppl the rtio test for bsolute convergence (Theorem 9.6.5). The following emple illustrtes how this works. Emple 3 Find the intervl of convergence nd rdius of convergence of the following power series. () k k (b) (c) k! k ( ) k k (d) k! 3 k (k + ) k= k= k= k=

251 Solution (). obtin 9.8 Mclurin nd Tlor Series; Power Series 663 Appling the rtio test for bsolute convergence to the given series, we ρ = lim u k+ k + u = lim k+ k k + k = lim = k + so the series converges bsolutel if ρ = < nd diverges if ρ = >. The test is inconclusive if = (i.e., if = or = ), which mens tht we will hve to investigte convergence t these vlues seprtel. At these vlues the series becomes k = = k= ( ) k = + + = k= both of which diverge; thus, the intervl of convergence for the given power series is (, ), nd the rdius of convergence is R =. Solution (b). Appling the rtio test for bsolute convergence to the given series, we obtin ρ = lim u k+ k + u = lim k+ k k + (k + )! k! k = lim k + k + = Since ρ< for ll, the series converges bsolutel for ll. Thus, the intervl of convergence is (, + ) nd the rdius of convergence is R =+. Solution (c). If =, then the rtio test for bsolute convergence ields ρ = lim u k+ k + u = lim (k + )! k+ k k + k! k = lim (k + ) =+ k + Therefore, the series diverges for ll nonzero vlues of. Thus, the intervl of convergence is the single vlue = nd the rdius of convergence is R =. Solution (d). Since ( ) k = ( ) k+ =, we obtin ρ = lim u k+ k + u = lim k+ k k + 3 k+ (k + ) 3k (k + ) k [ ( )] k + = lim k + 3 k + = ( ) + ( /k) lim = 3 k + + (/k) 3 The rtio test for bsolute convergence implies tht the series converges bsolutel if < 3 nd diverges if > 3. The rtio test fils to provide n informtion when =3, so the cses = 3 nd = 3 need seprte nlses. Substituting = 3 in the given series ields ( ) k ( 3) k ( ) k ( ) k 3 k = = 3 k (k + ) 3 k (k + ) k + k= k= k= which is the divergent hrmonic series Substituting = 3inthe 4 given series ields ( ) k 3 k 3 k (k + ) = ( ) k k + = k= which is the conditionll convergent lternting hrmonic series. Thus, the intervl of convergence for the given series is ( 3, 3] nd the rdius of convergence is R = 3. k=

252 664 Chpter 9 / Infinite Series POWER SERIES IN If is constnt, nd if is replced b in (3), then the resulting series hs the form c k ( ) k = c + c ( ) + c ( ) + +c k ( ) k + k= This is clled power series in. Some emples re ( ) k = + k + k= k= ( ) + ( ) 3 ( ) k ( + 3) k = ( + 3) + k! + ( + 3)! ( )3 4 + = ( + 3)3 3! + = 3 The first of these is power series in nd the second is power series in + 3. Note tht power series in is power series in in which =. More generll, the Tlor series f (k) ( ) ( ) k k! k= is power series in. The min result on convergence of power series in cn be obtined b substituting for in Theorem This leds to the following theorem theorem For power series c k ( ) k, ectl one of the following sttements is true: () The series converges onl for =. (b) The series converges bsolutel (nd hence converges) for ll rel vlues of. (c) The series converges bsolutel (nd hence converges) for ll in some finite open intervl ( R, + R) nd diverges if < R or > + R. At either of the vlues = R or = + R, the series m converge bsolutel, converge conditionll, or diverge, depending on the prticulr series. It follows from this theorem tht the set of vlues for which power series in converges is lws n intervl centered t = ; we cll this the intervl of convergence (Figure 9.8.). In prt () of Theorem the intervl of convergence reduces to the single vlue =, in which cse we s tht the series hs rdius of convergence R = ; in prt Diverges Diverges Rdius of convergence R = Converges Rdius of convergence R = + Figure 9.8. Diverges R Converges + R Diverges Rdius of convergence R

253 9.8 Mclurin nd Tlor Series; Power Series 665 (b) the intervl of convergence is infinite (the entire rel line), in which cse we s tht the series hs rdius of convergence R =+ ; nd in prt (c) the intervl etends between R nd + R, in which cse we s tht the series hs rdius of convergence R. Emple 4 Find the intervl of convergence nd rdius of convergence of the series ( 5) k k k= It will lws be wste of time to test for convergence t the endpoints of the intervl of convergence using the rtio test, since ρ will lws be t those points if lim k + u k+ u k eists. Eplin wh this must be so. Figure Solution. We ppl the rtio test for bsolute convergence. ρ = lim u k+ k + u = lim ( 5) k+ k k k + (k + ) ( 5) k [ ( ) ] k = lim 5 k + k + ( ) = 5 lim = 5 k + + (/k) Thus, the series converges bsolutel if 5 <, or < 5 <, or 4 <<6. The series diverges if <4or>6. To determine the convergence behvior t the endpoints = 4 nd = 6, we substitute these vlues in the given series. If = 6, the series becomes k= k k = k= k = which is convergent p-series (p = ). If = 4, the series becomes ( ) k = k= k Since this series converges bsolutel, the intervl of convergence for the given series is [4, 6]. The rdius of convergence is R = (Figure 9.8.3). Series diverges Series converges bsolutel 4 = 5 6 R = R = Series diverges FUNCTIONS DEFINED BY POWER SERIES If function f is epressed s power series on some intervl, then we s tht the power series represents f on tht intervl. For emple, we sw in Emple 4() of Section 9.3 tht = k if <, so this power series represents the function /( )on the intervl <<. k=

254 666 Chpter 9 / Infinite Series = J () = J () Generted b Mthemtic Figure TECHNOLOGY MASTERY Mn computer lgebr sstems hve the Bessel functions s prt of their librries. If ou hve CAS with Bessel functions, use it to generte the grphs in Figure Sometimes new functions ctull originte s power series, nd the properties of the functions re developed b working with their power series representtions. For emple, the functions ( ) k k J () = = k (k!) (!) (!) 6 + (4) 6 (3!) nd J () = k= k= ( ) k k+ k+ (k!)(k + )! = 3 3 (!)(!) (!)(3!) (5) which re clled Bessel functions in honor of the Germn mthemticin nd stronomer Friedrich Wilhelm Bessel ( ), rise nturll in the stud of plnetr motion nd in vrious problems tht involve het flow. To find the domins of these functions, we must determine where their defining power series converge. For emple, in the cse of J () we hve ρ = lim u k+ k + u = lim (k+) k k + (k+) [(k + )!] k (k!) k = lim k + 4(k + ) = < so the series converges for ll ; tht is, the domin of J () is (, + ). We leve it s n eercise (Eercise 6) to show tht the power series for J () lso converges for ll. Computer-generted grphs of J () nd J () re shown in Figure QUICK CHECK EXERCISES 9.8 (See pge 668 for nswers.). If f hs derivtives of ll orders t, then the Tlor series for f bout = is defined to be. Since lim k + k= k+ k+ k k = the rdius of convergence for the infinite series k= k k is. 3. Since lim (3 k+ k+ )/(k + )! k + (3 k k )/k! = lim 3 k + k + = the intervl of convergence for the series k= (3k /k!) k is. 4. () Since ( 4) k+ / k + lim ( 4) k / k = lim k + k + = 4 k k + ( 4) the rdius of convergence for the infinite series k= ( / k)( 4) k is. (b) When = 3, k= k ( 4) k = k= Does this series converge or diverge? (c) When = 5, k= k ( 4) k = k ( ) k k= Does this series converge or diverge? k (d) The intervl of convergence for the infinite series k= ( / k)( 4) k is.

255 9.8 Mclurin nd Tlor Series; Power Series 667 EXERCISE SET 9.8 Grphing Utilit C CAS Use sigm nottion to write the Mclurin series for the function.. e. e 3. cos π 4. sin π 5. ln( + ) cosh + 8. sinh 9. sin. e 8 Use sigm nottion to write the Tlor series bout = for the function.. e ; =. e ; = ln 3. ; = 4. + ; = 3 5. sin π; = 6. cos ; = π 7. ln ; = 8. ln ; = e 9 Find the intervl of convergence of the power series, nd find fmilir function tht is represented b the power series on tht intervl ( ) k k k +. + ( ) + ( ) + +( ) k +. ( + 3) + ( + 3) ( + 3) 3 + +( ) k ( + 3) k Suppose tht the function f is represented b the power series f() = k + +( )k 8 + k () Find the domin of f. (b) Find f() nd f(). 4. Suppose tht the function f is represented b the power series f() = 5 ( 5) ( 5) () Find the domin of f. (b) Find f(3) nd f(6). 5 8 True Flse Determine whether the sttement is true or flse. Eplin our nswer. 5. If power series in converges conditionll t = 3, then the series converges if < 3 nd diverges if > The rtio test is often useful to determine convergence t the endpoints of the intervl of convergence of power series. 7. The Mclurin series for polnomil function hs rdius of convergence +. k 8. The series converges if <. k! k= 9 5 Find the rdius of convergence nd the intervl of convergence. k k k ( ) k k 3. k + k! k= k= k= k! 5 k k 3. k k 33. k k 34. ln k k= k= k= k ( ) k k k(k + ) k + k= k= k k ( ) k k 37. ( ) 38. k= k (k)! k= 3 k 39. k! k 4. ( ) k+ k k(ln k) k= k= k ( ) k k+ + k (k + )! k= k= ( ) 3 k 43. ( + 5) k ( 3) k k k= k= k+ ( + )k k ( 4)k 45. ( ) 46. ( ) k (k + ) k= k= k ( + )k+ (k + )! 47. ( ) 48. ( ) k k + 4 k 3 k= k= π k ( ) k ( 3) k (k + )! 4 k k= k= 5. Use the root test to find the intervl of convergence of k (ln k) k k= 5. Find the domin of the function 3 5 (k ) f() = k (k )! k= 53. Show tht the series! + 4! 3 6! + is the Mclurin series{ for the function cos, f() = cosh, < [Hint: Use the Mclurin series for cos nd cosh to obtin series for cos, where, nd cosh, where.] FOCUS ON CONCEPTS 54. If function f is represented b power series on n intervl, then the grphs of the prtil sums cn be used s pproimtions to the grph of f. () Use grphing utilit to generte the grph of /( ) together with the grphs of the first four prtil sums of its Mclurin series over the intervl (, ). (cont.)

256 668 Chpter 9 / Infinite Series (b) In generl terms, where re the grphs of the prtil sums the most ccurte? 55. Prove: () If f is n even function, then ll odd powers of in its Mclurin series hve coefficient. (b) If f is n odd function, then ll even powers of in its Mclurin series hve coefficient. 56. Suppose tht the power series c k ( ) k hs rdius of convergence R nd p is nonzero constnt. Wht cn ou s bout the rdius of convergence of the power series pc k ( ) k? Eplin our resoning. [Hint: See Theorem ] 57. Suppose tht the power series c k ( ) k hsfinite rdius of convergence R, nd the power series dk ( ) k hs rdius of convergence of +. Wht cn ou s bout the rdius of convergence of (ck + d k )( ) k? Eplin our resoning. 58. Suppose tht the power series c k ( ) k hsfinite rdius of convergence R nd the power series dk ( ) k hs finite rdius of convergence R. Wht cn ou s bout the rdius of convergence of (ck + d k )( ) k? Eplin our resoning. [Hint: The cse R = R requires specil ttention.] 59. Show tht if p is positive integer, then the power series (pk)! (k!) p k k= hs rdius of convergence of /p p. 6. Show tht if p nd q re positive integers, then the power series (k + p)! k!(k + q)! k k= hs rdius of convergence of +. C 6. Show tht the power series representtion of the Bessel function J () converges for ll [Formul (5)]. 6. Approimte the vlues of the Bessel functions J () nd J () t =, ech to four deciml-plce ccurc. 63. If the constnt p in the generl p-series is replced b vrible for >, then the resulting function is clled the Riemnn zet function nd is denoted b ζ() = () Let s n be the nth prtil sum of the series for ζ(3.7). Find n such tht s n pproimtes ζ(3.7) to two decimlplce ccurc, nd clculte s n using this vlue of n. [Hint: Use the right inequlit in Eercise 36(b) of Section 9.4 with f() = / 3.7.] (b) Determine whether our CAS cn evlute the Riemnn zet function directl. If so, compre the vlue produced b the CAS to the vlue of s n obtined in prt (). 64. Prove: If lim k + c k /k = L, where L =, then /L is the rdius of convergence of the power series k= c k k. 65. Prove: If the power series k= c k k hs rdius of convergence R, then the series k= c k k hs rdius of convergence R. 66. Prove: If the intervl of convergence of the series k= c k( ) k is ( R, + R], then the series converges conditionll t + R. 67. Writing The sine function cn be defined geometricll from the unit circle or nlticll from its Mclurin series. Discuss the dvntges of ech representtion with regrd to providing informtion bout the sine function. k= k QUICK CHECK ANSWERS 9.8. f (k) ( ) ( ) k. k! 3. (, + ) 4. () (b) converges (c) diverges (d) [3, 5) 9.9 CONVERGENCE OF TAYLOR SERIES In this section we will investigte when Tlor series for function converges to tht function on some intervl, nd we will consider how Tlor series cn be used to pproimte vlues of trigonometric, eponentil, nd logrithmic functions. THE CONVERGENCE PROBLEM FOR TAYLOR SERIES Recll tht the nth Tlor polnomil for function f bout = hs the propert tht its vlue nd the vlues of its first n derivtives mtch those of f t. As n increses,

257 9.9 Convergence of Tlor Series 669 more nd more derivtives mtch up, so it is resonble to hope tht for vlues of ner the vlues of the Tlor polnomils might converge to the vlue of f(); tht is, n f (k) ( ) f() = lim ( ) k () n + k! k= However, the nth Tlor polnomil for f is the nth prtil sum of the Tlor series for f, so () is equivlent to stting tht the Tlor series for f converges t, nd its sum is f(). Thus, we re led to consider the following problem. Problem 9.9. is concerned not onl with whether the Tlor series of function f converges, but lso whether it converges to the function f itself. Indeed, it is possible for Tlor series of function f to converge to vlues different from f() for certin vlues of (Eercise 4) problem Given function f tht hs derivtives of ll orders t =, determine whether there is n open intervl contining such tht f() is the sum of its Tlor series bout = t ech point in the intervl; tht is, f (k) ( ) f() = ( ) k () k! for ll vlues of in the intervl. k= One w to show tht () holds is to show tht [ n f() lim n + k= ] f (k) ( ) ( ) k = k! However, the difference ppering on the left side of this eqution is the nth reminder for the Tlor series [Formul () of Section 9.7]. Thus, we hve the following result theorem The equlit f() = k= f (k) ( ) ( ) k k! holds t point if nd onl if lim n + R n() =. ESTIMATING THE nth REMAINDER It is reltivel rre tht one cn prove directl tht R n () sn +. Usull, this is proved indirectl b finding pproprite bounds on R n () nd ppling the Squeezing Theorem for Sequences. The Reminder Estimtion Theorem (Theorem 9.7.4) provides useful bound for this purpose. Recll tht this theorem sserts tht if M is n upper bound for f (n+) () on n intervl contining, then R n () M (n + )! n+ (3) for ll in tht intervl. The following emple illustrtes how the Reminder Estimtion Theorem is pplied. Emple is, cos = Show tht the Mclurin series for cos converges to cos for ll ; tht ( ) k k (k)! =! + 4 4! 6 + ( <<+ ) 6! k=

258 67 Chpter 9 / Infinite Series Solution. From Theorem 9.9. we must show tht R n () for ll s n +. For this purpose let f() = cos, so tht for ll we hve f (n+) () =±cos or f (n+) () =±sin The method of Emple cn be esil modified to prove tht the Tlor series for sin nd cos bout n point = converge to sin nd cos, respectivel, for ll (Eercises nd ). For reference, some of the most importnt Mclurin series re listed in Tble 9.9. t the end of this section. In ll cses we hve f (n+) (), so we cn ppl (3) with M = nd = to conclude tht R n () n+ (4) (n + )! However, it follows from Formul (5) of Section 9. with n + in plce of n nd in plce of tht n+ lim n + (n + )! = (5) Using this result nd the Squeezing Theorem for Sequences (Theorem 9..5), it follows from (4) tht R n () nd hence tht R n () sn + (Theorem 9..6). Since this is true for ll, we hve proved tht the Mclurin series for cos converges to cos for ll. This is illustrted in Figure 9.9., where we cn see how successive prtil sums pproimte the cosine curve more nd more closel. p 4 p 8 p p 6 = cos n p n = ( ) k k (k)! k= Figure 9.9. p p 6 p p 4 p 8 APPROXIMATING TRIGONOMETRIC FUNCTIONS In generl, to pproimte the vlue of function f t point using Tlor series, there re two bsic questions tht must be nswered: About wht point should the Tlor series be epnded? How mn terms in the series should be used to chieve the desired ccurc? In response to the first question, needs to be point t which the derivtives of f cn be evluted esil, since these vlues re needed for the coefficients in the Tlor series. Furthermore, if the function f is being evluted t, then should be chosen s close s possible to, since Tlor series tend to converge more rpidl ner. For emple, to pproimte sin 3 (= π/6 rdins), it would be resonble to tke =, since π/6 is close to nd the derivtives of sin re es to evlute t. On the other hnd, to pproimte sin 85 (= 7π/36 rdins), it would be more nturl to tke = π/, since 7π/36 is close to π/ nd the derivtives of sin re es to evlute t π/. In response to the second question posed bove, the number of terms required to chieve specific ccurc needs to be determined on problem-b-problem bsis. The net emple gives two methods for doing this. Use the Mclurin series for sin to pproimte sin 3 to five deciml- Emple plce ccurc.

259 Solution. 9.9 Convergence of Tlor Series 67 In the Mclurin series sin = ( ) k k+ (k + )! = 3 3! + 5 5! 7 + (6) 7! k= the ngle is ssumed to be in rdins (becuse the differentition formuls for the trigonometric functions were derived with this ssumption). Since 3 = π/6 rdins, it follows from (6) tht sin 3 = sin π 6 = ( π 6 ) (π /6) 3 3! + (π /6) 5 5! (π /6) 7 7! + (7) We must now determine how mn terms in the series re required to chieve five decimlplce ccurc. We will consider two possible pproches, one using the Reminder Estimtion Theorem (Theorem 9.7.4) nd the other using the fct tht (7) stisfies the hpotheses of the lternting series test (Theorem 9.6.). Method. (The Reminder Estimtion Theorem) Since we wnt to chieve five deciml-plce ccurc, our gol is to choose n so tht the bsolute vlue of the nth reminder t = π/6 does not eceed.5 = 5 6 ; tht is, ( Rn π ).5 (8) 6 However, if we let f() = sin, then f (n+) () is either ± sin or ± cos, nd in either cse f (n+) () for ll. Thus, it follows from the Reminder Estimtion Theorem with M =, =, nd = π/6 tht ( π ) (π/6) n+ R n 6 (n + )! Thus, we cn stisf (8) b choosing n so tht (π/6) n+ (n + )!.5 With the help of clculting utilit ou cn verif tht the smllest vlue of n tht meets this criterion is n = 3. Thus, to chieve five deciml-plce ccurc we need onl keep terms up to the third power in (7). This ields ( π ) sin 3 (π /6) (9) 6 3! (verif). As check, clcultor gives sin , which grees with (9) when rounded to five deciml plces. Method. (The Alternting Series Test) We leve it for ou to check tht (7) stisfies the hpotheses of the lternting series test (Theorem 9.6.). Let s n denote the sum of the terms in (7) up to nd including the nth power of π/6. Since the eponents in the series re odd integers, the integer n must be odd, nd the eponent of the first term not included in the sum s n must be n +. Thus, it follows from prt (b) of Theorem 9.6. tht sin 3 s n < (π /6) n+ (n + )! This mens tht for five deciml-plce ccurc we must look for the first positive odd integer n such tht (π/6) n+.5 (n + )!

260 67 Chpter 9 / Infinite Series With the help of clculting utilit ou cn verif tht the smllest vlue of n tht meets this criterion is n = 3. This grees with the result obtined bove using the Reminder Estimtion Theorem nd hence leds to pproimtion (9) s before. ROUNDOFF AND TRUNCATION ERROR There re two tpes of errors tht occur when computing with series. The first, clled trunction error, is the error tht results when series is pproimted b prtil sum; nd the second, clled roundoff error, is the error tht rises from pproimtions in numericl computtions. For emple, in our derivtion of (9) we took n = 3 to keep the trunction error below.5. However, to evlute the prtil sum we hd to pproimte π, thereb introducing roundoff error. Hd we not eercised some cre in choosing this pproimtion, the roundoff error could esil hve degrded the finl result. Methods for estimting nd controlling roundoff error re studied in brnch of mthemtics clled numericl nlsis. However, s rule of thumb, to chieve n deciml-plce ccurc in finl result, ll intermedite clcultions must be ccurte to t lest n + deciml plces. Thus, in (9) t lest si deciml-plce ccurc in π is required to chieve the five deciml-plce ccurc in the finl numericl result. As prcticl mtter, good working procedure is to perform ll intermedite computtions with the mimum number of digits tht our clculting utilit cn hndle nd then round t the end. APPROXIMATING EXPONENTIAL FUNCTIONS Emple 3 Show tht the Mclurin series for e converges to e for ll ; tht is, e k = k! = + +! + 3 k ( <<+ ) 3! k! k= Solution. Let f() = e, so tht f (n+) () = e We wnt to show tht R n () sn + for ll in the intervl <<+. However, it will be helpful here to consider the cses nd > seprtel. If, then we will tke the intervl in the Reminder Estimtion Theorem (Theorem 9.7.4) to be [,], nd if >, then we will tke it to be [,]. Since f (n+) () = e is n incresing function, it follows tht if c is in the intervl [,], then nd if c is in the intervl [,], then f (n+) (c) f (n+) () =e = f (n+) (c) f (n+) () =e Thus, we cn ppl Theorem with M = in the cse where nd with M = e in the cse where >. This ields R n () n+ (n + )! if R n () e n+ if > (n + )! Thus, in both cses it follows from (5) nd the Squeezing Theorem for Sequences tht R n () sn +, which in turn implies tht R n () sn +. Since this is true for ll, we hve proved tht the Mclurin series for e converges to e for ll.

261 9.9 Convergence of Tlor Series 673 Since the Mclurin series for e converges to e for ll, we cn use prtil sums of the Mclurin series to pproimte powers of e to rbitrr precision. Recll tht in Emple 7 of Section 9.7 we were ble to use the Reminder Estimtion Theorem to determine tht evluting the ninth Mclurin polnomil for e t = ields n pproimtion for e with five deciml-plce ccurc: e + +! + 3! + 4! + 5! + 6! + 7! + 8! + 9!.788 APPROXIMATING LOGARITHMS The Mclurin series In Emple of Section 9.6, we stted without proof tht ln = This result cn be obtined b letting = in (), but s indicted in the tet discussion, this series converges too slowl to be of prcticl use. Jmes Gregor ( ) Scottish mthemticin nd stronomer. Gregor, the son of minister, ws fmous in his time s the inventor of the Gregorin reflecting telescope, so nmed in his honor. Although he is not generll rnked with the gret mthemticins, much of his work relting to clculus ws studied b Leibniz nd Newton nd undoubtedl influenced some of their discoveries. There is mnuscript, discovered posthumousl, which shows tht Gregor hd nticipted Tlor series well before Tlor. [Imge: File:Jmes_Gregor.jpeg] ln( + ) = ( < ) () 4 is the strting point for the pproimtion of nturl logrithms. Unfortuntel, the usefulness of this series is limited becuse of its slow convergence nd the restriction <. However, if we replce b in this series, we obtin ln( ) = ( <) () nd on subtrcting () from () we obtin ( ) ) + ln = ( ( <<) () Series (), first obtined b Jmes Gregor in 668, cn be used to compute the nturl logrithm of n positive number b letting = + or, equivlentl, = + (3) nd noting tht <<. For emple, to compute ln we let = in (3), which ields =. Substituting this vlue in () gives 3 [ ( 3 ( 5 ( 7 ] ln = 3) 3) 3) (4) In Eercise 9 we will sk ou to show tht five deciml-plce ccurc cn be chieved using the prtil sum with terms up to nd including the 3th power of. Thus, to five 3 deciml-plce ccurc [ ( 3 ( 5 ( 7 ( 3 ] ln 3) 3) 3) 3) (verif). As check, clcultor gives ln , which grees with the preceding pproimtion when rounded to five deciml plces. APPROXIMATING π In the net section we will show tht tn = ( ) (5) Letting =, we obtin π 4 = tn =

262 674 Chpter 9 / Infinite Series or π = 4 [ + + ] This fmous series, obtined b Leibniz in 674, converges too slowl to be of computtionl vlue. A more prcticl procedure for pproimting π uses the identit π 4 = tn + tn (6) 3 which ws derived in Eercise 6 of Section.4. B using this identit nd series (5) to pproimte tn nd tn, the vlue of π cn be pproimted efficientl to n 3 degree of ccurc. Let f() = ( + ) m. Verif tht f() = f () = m f () = m(m ) f () = m(m )(m ). f (k) () = m(m ) (m k + ) BINOMIAL SERIES If m is rel number, then the Mclurin series for ( + ) m is clled the binomil series; it is given b m(m ) + m + m(m )(m ) + 3 m(m ) (m k + ) + + k +! 3! k! In the cse where m is nonnegtive integer, the function f() = ( + ) m is polnomil of degree m,so f (m+) () = f (m+) () = f (m+3) () = = nd the binomil series reduces to the fmilir binomil epnsion ( + ) m m(m ) = + m + m(m )(m ) m! 3! which is vlid for <<+. It cn be proved tht if m is not nonnegtive integer, then the binomil series converges to ( + ) m if <. Thus, for such vlues of ( + ) m = + m + or in sigm nottion, m(m ) + +! m(m ) (m k + ) k + (7) k! ( + ) m = + k= m(m ) (m k + ) k if < (8) k! Emple 4 Find binomil series for () ( + ) (b) + Solution (). Since the generl term of the binomil series is complicted, ou m find it helpful to write out some of the beginning terms of the series, s in Formul (7), to see developing ptterns. Substituting m = in this formul ields ( + ) = ( + ) = + ( ) + ( )( 3)! + ( )( 3)( 4) 3! = + 3!! 4! 3! 3 + 5! 4! 4 = = ( ) k (k + ) k k= 3 + ( )( 3)( 4)( 5) 4 + 4!

263 9.9 Convergence of Tlor Series 675 p 3 () 4 3 = ( + ) Solution (b). Substituting m = in (7) ields = ( )( + + ) ( +! = + 3! ! 3 + k 3 5 (k ) = + ( ) k k k! k= )( )( ) 3 + 3! p 3 () = = + p 3 () p 3 () = Figure 9.9. Figure 9.9. shows the grphs of the functions in Emple 4 compred to their thirddegree Mclurin polnomils. SOME IMPORTANT MACLAURIN SERIES For reference, Tble 9.9. lists the Mclurin series for some of the most importnt functions, together with specifiction of the intervls over which the Mclurin series converge to those functions. Some of these results re derived in the eercises nd others will be derived in the net section using some specil techniques tht we will develop. Tble 9.9. some importnt mclurin series mclurin series intervl of convergence = k= k = < < + = k= ( ) k k = < < e = k = k!! 3! 4! k= < < + sin = ( ) k k+ = (k + )! 3! 5! 7! k= < < + cos = ( ) k k = (k)!! 4! 6! k= < < + ln ( + ) = ( ) k+ k = k 3 4 k= < tn = ( ) k k+ = k k= 3 5 sinh = k+ = (k + )! 3! 5! 7! k= < < + cosh = k = (k)!! 4! 6! k= < < + ( + ) m m(m )... (m k + ) = + k < < * k! (m,,,...) k= * The behvior t the endpoints depends on m: For m > the series converges bsolutel t both endpoints; for m the series diverges t both endpoints; nd for < m < the series converges conditionll t = nd diverges t =.

264 676 Chpter 9 / Infinite Series QUICK CHECK EXERCISES 9.9 (See pge 677 for nswers.). cos =. e = k= k= 3. ln( + ) = converges to ( + ) m if <. for in the intervl. k= 4. If m is rel number but not nonnegtive integer, the binomil series + k= EXERCISE SET 9.9 Grphing Utilit C CAS. Use the Reminder Estimtion Theorem nd the method of Emple to prove tht the Tlor series for sin bout = π/4 converges to sin for ll.. Use the Reminder Estimtion Theorem nd the method of Emple 3 to prove tht the Tlor series for e bout = converges to e for ll. 3 Approimte the specified function vlue s indicted nd check our work b compring our nswer to the function vlue produced directl b our clculting utilit. 3. Approimte sin 4 to five deciml-plce ccurc using both of the methods given in Emple. 4. Approimte cos 3 to three deciml-plce ccurc using both of the methods given in Emple. 5. Approimte cos. to five deciml-plce ccurc using the Mclurin series for cos. 6. Approimte tn. to three deciml-plce ccurc using the Mclurin series for tn. 7. Approimte sin 85 to four deciml-plce ccurc using n pproprite Tlor series. 8. Approimte cos( 75 ) to four deciml-plce ccurc using Tlor series. 9. Approimte sinh.5 to three deciml-plce ccurc using the Mclurin series for sinh.. Approimte cosh. to three deciml-plce ccurc using the Mclurin series for cosh.. () Use Formul () in the tet to find series tht converges to ln.5. (b) Approimte ln.5 using the first two terms of the series. Round our nswer to three deciml plces, nd compre the result to tht produced directl b our clculting utilit.. () Use Formul () to find series tht converges to ln 3. (b) Approimte ln 3 using the first two terms of the series. Round our nswer to three deciml plces, nd compre the result to tht produced directl b our clculting utilit. FOCUS ON CONCEPTS 3. () Use the Mclurin series for tn to pproimte tn nd tn to three deciml-plce ccurc. 3 (b) Use the results in prt () nd Formul (6) to pproimte π. (c) Would ou be willing to gurntee tht our nswer in prt (b) is ccurte to three deciml plces? Eplin our resoning. (d) Compre our nswer in prt (b) to tht produced b our clculting utilit. 4. The purpose of this eercise is to show tht the Tlor series of function f m possibl converge to vlue different from f() for certin vlues of. Let { e /, = f() =, = () Use the definition of derivtive to show tht f () =. (b) With some difficult it cn be shown tht if n, then f (n) () =. Accepting this fct, show tht the Mclurin series of f converges for ll but converges to f() onl t =. 5. () Find n upper bound on the error tht cn result if cos is pproimted b ( /!) + ( 4 /4!) over the intervl [.,.]. (b) Check our nswer in prt () b grphing ( ) cos! + 4 4! over the intervl. 6. () Find n upper bound on the error tht cn result if ln( + ) is pproimted b over the intervl [.,.]. (b) Check our nswer in prt () b grphing over the intervl. ln( + )

265 7. Use Formul (7) for the binomil series to obtin the Mclurin series for () 3 (b) + (c) + ( + ) If m is n rel number, nd k is nonnegtive integer, then we define the binomil coefficient ( ) ( ) m m b the formuls = nd k ( ) m m(m )(m ) (m k + ) = k k! for k. Epress Formul (7) in the tet in terms of binomil coefficients. 9. In this eercise we will use the Reminder Estimtion Theorem to determine the number of terms tht re required in Formul (4) to pproimte ln to five deciml-plce ccurc. For this purpose let f() = ln + = ln( + ) ln( ) ( <<) () Show tht [ ] ( ) f (n+) n () = n! ( + ) + n+ ( ) n+ (b) Use the tringle inequlit [Theorem..4(d)] to show tht [ ] f (n+) () n! ( + ) + n+ ( ) n+ (c) Since we wnt to chieve five deciml-plce ccurc, our gol is to choose n so tht the bsolute vlue of the nth reminder t = does not eceed the vlue 3.5 =.5 5 ; tht is, ( R ) n.5. 3 Use the Reminder Estimtion Theorem to show tht this condition will be stisfied if n is chosen so tht ( ) M n+.5 (n + )! 3 where f (n+) () M on the intervl [ ], 3. (d) Use the result in prt (b) [ to show tht ] M cn be tken s M = n! + ( ) n+ 3 C 9.9 Convergence of Tlor Series 677 (e) Use the results in prts (c) nd (d) to show tht five deciml-plce ccurc will be chieved if n stisfies [ ( ) n+ ( ) ] n+ +.5 n + 3 nd then show tht the smllest vlue of n tht stisfies this condition is n = 3.. Use Formul () nd the method of Eercise 9 to pproimte ln ( ) 5 3 to five deciml-plce ccurc. Then check our work b compring our nswer to tht produced directl b our clculting utilit.. Prove: The Tlor series for cos bout n vlue = converges to cos for ll.. Prove: The Tlor series for sin bout n vlue = converges to sin for ll. 3. Reserch hs shown tht the proportion p of the popultion with IQs (intelligence quotients) between α nd β is pproimtel β p = 6 e ( 6 ) d π α Use the first three terms of n pproprite Mclurin series to estimte the proportion of the popultion tht hs IQs between nd. 4. () In 76 the British stronomer nd mthemticin John Mchin discovered the following formul for π/4, clled Mchin s formul: π 4 = 4 tn 5 tn 39 Use CAS to pproimte π/4 using Mchin s formul to 5 deciml plces. (b) In 94 the brillint Indin mthemticin Srinivs Rmnujn (887 9) showed tht 8 π = 98 k= (4k)!(3 + 6,39k) (k!) k Use CAS to compute the first four prtil sums in Rmnujn s formul. QUICK CHECK ANSWERS 9.9. ( ) k k (k)!. k k! k+ k 3. ( ) k ; (, ] 4. m(m ) (m k + ) k ; k!

266 678 Chpter 9 / Infinite Series 9. DIFFERENTIATING AND INTEGRATING POWER SERIES; MODELING WITH TAYLOR SERIES In this section we will discuss methods for finding power series for derivtives nd integrls of functions, nd we will discuss some prcticl methods for finding Tlor series tht cn be used in situtions where it is difficult or impossible to find the series directl. DIFFERENTIATING POWER SERIES We begin b considering the following problem. 9.. problem Suppose tht function f is represented b power series on n open intervl. How cn we use the power series to find the derivtive of f on tht intervl? The solution to this problem cn be motivted b considering the Mclurin series for sin : sin = 3 3! + 5 5! 7 + ( <<+ ) 7! Of course, we lred know tht the derivtive of sin is cos ; however, we re concerned here with using the Mclurin series to deduce this. The solution is es ll we need to do is differentite the Mclurin series term b term nd observe tht the resulting series is the Mclurin series for cos : ] d [ 3 d 3! + 5 5! 7 7! + = 3 3! ! 76 7! + Here is nother emple. d d [e ]= d d [ + +! + 3 3! + 4 4! + =! + 4 4! 6 6! + =cos = +! + 3 3! = + + 4!! + 3 3! + =e The preceding computtions suggest tht if function f is represented b power series on n open intervl, then power series representtion of f on tht intervl cn be obtined b differentiting the power series for f term b term. This is stted more precisel in the following theorem, which we give without proof. 9.. theorem (Differentition of Power Series) Suppose tht function f is represented b power series in tht hs nonzero rdius of convergence R; tht is, f() = c k ( ) k ( R<< + R) Then: () (b) k= The function f is differentible on the intervl ( R, + R). If the power series representtion for f is differentited term b term, then the resulting series hs rdius of convergence R nd converges to f on the intervl ( R, + R); tht is, f () = k= ] d d [c k( ) k ] ( R<< + R)

267 9. Differentiting nd Integrting Power Series; Modeling with Tlor Series 679 This theorem hs n importnt impliction bout the differentibilit of functions tht re represented b power series. According to the theorem, the power series for f hs the sme rdius of convergence s the power series for f, nd this mens tht the theorem cn be pplied to f s well s f. However, if we do this, then we conclude tht f is differentible on the intervl ( R, + R), nd the power series for f hs the sme rdius of convergence s the power series for f nd f. We cn now repet this process d infinitum, ppling the theorem successivel to f, f,...,f (n),...to conclude tht f hs derivtives of ll orders on the intervl ( R, + R). Thus, we hve estblished the following result theorem If function f cn be represented b power series in with nonzero rdius of convergence R, then f hs derivtives of ll orders on the intervl ( R, + R). In short, it is onl the most well-behved functions tht cn be represented b power series; tht is, if function f does not possess derivtives of ll orders on n intervl ( R, + R), then it cnnot be represented b power series in on tht intervl. Emple the power series In Section 9.8, we showed tht the Bessel function J (), represented b J () = k= ( ) k k k (k!) () hs rdius of convergence + [see Formul (4) of tht section nd the relted discussion]. Thus, J () hs derivtives of ll orders on the intervl (, + ), nd these cn be obtined b differentiting the series term b term. For emple, if we write () s J () = + k= ( ) k k k (k!) See Eercise 45 for reltionship between J () nd J (). nd differentite term b term, we obtin J () = ( ) k (k) k ( ) k k = k (k!) k k!(k )! k= k= REMARK The computtions in this emple use some techniques tht re worth noting. First, when power series is epressed in sigm nottion, the formul for the generl term of the series will often not be of form tht cn be used for differentiting the constnt term. Thus, if the series hs nonzero constnt term, s here, it is usull good ide to split it off from the summtion before differentiting. Second, observe how we simplified the finl formul b cnceling the fctor k from one of the fctorils in the denomintor. This is stndrd simplifiction technique. INTEGRATING POWER SERIES Since the derivtive of function tht is represented b power series cn be obtined b differentiting the series term b term, it should not be surprising tht n ntiderivtive of function represented b power series cn be obtined b integrting the series term b term. For emple, we know tht sin is n ntiderivtive of cos. Here is how this result

268 68 Chpter 9 / Infinite Series cn be obtined b integrting the Mclurin series for cos term b term: ] cos d= [! + 4 4! 6 6! + d ] = [ 3 3(!) + 5 5(4!) 7 7(6!) + + C ] = [ 3 3! + 5 5! 7 7! + + C = sin + C The sme ide pplies to definite integrls. For emple, b direct integrtion we hve d + = tn ] nd we will show lter in this section tht Thus, = tn tn = π 4 = π 4 π 4 = () 7 d + = Here is how this result cn be obtined b integrting the Mclurin series for /( + ) term b term (see Tble 9.9.): d + = [ ]d = ] = The preceding computtions re justified b the following theorem, which we give without proof. Theorems 9.. nd 9..4 tell us how to use power series representtion of function f to produce power series representtions of f () nd f()d tht hve the sme rdius of convergence s f. However, the intervls of convergence for these series m not be the sme becuse their convergence behvior m differ t the endpoints of the intervl. (See Eercises 5 nd 6.) 9..4 theorem (Integrtion of Power Series) Suppose tht function f is represented b power series in tht hs nonzero rdius of convergence R; tht is, f() = c k ( ) k ( R<< + R) () (b) k= If the power series representtion of f is integrted term b term, then the resulting series hs rdius of convergence R nd converges to n ntiderivtive for f() on the intervl ( R, + R); tht is, [ ] ck f()d = k + ( ) k+ + C ( R<< + R) k= If α nd β re points in the intervl ( R, + R), nd if the power series representtion of f is integrted term b term from α to β, then the resulting series converges bsolutel on the intervl ( R, + R) nd β [ β ] f()d = c k ( ) k d α k= α

269 9. Differentiting nd Integrting Power Series; Modeling with Tlor Series 68 POWER SERIES REPRESENTATIONS MUST BE TAYLOR SERIES For mn functions it is difficult or impossible to find the derivtives tht re required to obtin Tlor series. For emple, to find the Mclurin series for /( + ) directl would require some tedious derivtive computtions (tr it). A more prcticl pproch is to substitute for in the geometric series to obtin = ( <<) + = However, there re two questions of concern with this procedure: Where does the power series tht we obtined for /( + ) ctull converge to /( + )? How do we know tht the power series we hve obtined is ctull the Mclurin series for /( + )? The first question is es to resolve. Since the geometric series converges to /( ) if <, the second series will converge to /( + ) if < or <. However, this is true if nd onl if <, so the power series we obtined for the function /( + ) converges to this function if <<. The second question is more difficult to nswer nd leds us to the following generl problem problem Suppose tht function f is represented b power series in tht hs nonzero rdius of convergence. Wht reltionship eists between the given power series nd the Tlor series for f bout =? The nswer is tht the re the sme; nd here is the theorem tht proves it. Theorem 9..6 tells us tht no mtter how we rrive t power series representtion of function f,beitb substitution, b differentition, b integrtion, or b some lgebric process, tht series will be the Tlor series for f bout =, provided the series converges to f on some open intervl contining theorem If function f is represented b power series in on some open intervl contining, then tht power series is the Tlor series for f bout =. proof Suppose tht f() = c + c ( ) + c ( ) + +c k ( ) k + for ll in some open intervl contining. To prove tht this is the Tlor series for f bout =, we must show tht c k = f (k) ( ) for k =,,, 3,... k! However, the ssumption tht the series converges to f() on n open intervl contining ensures tht it hs nonzero rdius of convergence R; hence we cn differentite term

270 68 Chpter 9 / Infinite Series b term in ccordnce with Theorem 9... Thus, f() = c + c ( ) + c ( ) + c 3 ( ) 3 + c 4 ( ) 4 + f () = c + c ( ) + 3c 3 ( ) + 4c 4 ( ) 3 + f () =!c + (3 )c 3 ( ) + (4 3)c 4 ( ) + f () = 3!c 3 + (4 3 )c 4 ( ) +. On substituting =, ll the powers of drop out, leving from which we obtin f( ) = c, f ( ) = c, f ( ) =!c, f ( ) = 3!c 3,... c = f( ), c = f ( ), c = f ( ), c 3 = f ( ),...! 3! which shows tht the coefficients c, c, c, c 3,...re precisel the coefficients in the Tlor series bout for f(). SOME PRACTICAL WAYS TO FIND TAYLOR SERIES Emple Find Tlor series for the given functions bout the given. () e, = (b) ln, = (c), = Solution (). The simplest w to find the Mclurin series for e is to substitute for in the Mclurin series to obtin e = + +! + 3 3! (3) 4! e = + 4! 6 3! + 8 4! Since (3) converges for ll vlues of, so will the series for e. Solution (b). Tble 9.9.: We begin with the Mclurin series for ln( + ), which cn be found in ln( + ) = ( < ) 4 Substituting for in this series gives ( ) ( )3 ( )4 ln( +[ ]) = ln = ( ) + + (4) 3 4 Since the originl series converges when <, the intervl of convergence for (4) will be < or, equivlentl, <. Solution (c). Since / is the derivtive of ln, we cn differentite the series for ln found in (b) to obtin ( ) 3( ) 4( )3 = = ( ) + ( ) ( ) 3 + (5)

271 9. Differentiting nd Integrting Power Series; Modeling with Tlor Series 683 B Theorem 9.., we know tht the rdius of convergence for (5) is the sme s tht for (4), which is R =. Thus the intervl of convergence for (5) must be t lest <<. Since the behviors of (4) nd (5) m differ t the endpoints = nd =, those must be checked seprtel. When =, (5) becomes ( ) + ( ) ( ) 3 + = which diverges b the divergence test. Similrl, when =, (5) becomes = + + which lso diverges b the divergence test. Thus the intervl of convergence for (5) is <<. Emple 3 Find the Mclurin series for tn. Solution. It would be tedious to find the Mclurin series directl. A better pproch is to strt with the formul + d = tn + C nd integrte the Mclurin series + = ( <<) term b term. This ields tn + C = + d = [ ]d or ] tn = [ C The constnt of integrtion cn be evluted b substituting = nd using the condition tn =. This gives C =, so tht tn = ( <<) (6) 9 REMARK Observe tht neither Theorem 9.. nor Theorem 9..3 ddresses wht hppens t the endpoints of the intervl of convergence. However, it cn be proved tht if the Tlor series for f bout = converges to f() for ll in the intervl ( R, + R), nd if the Tlor series converges t the right endpoint + R, then the vlue tht it converges to t tht point is the limit of f() s + R from the left; nd if the Tlor series converges t the left endpoint R, then the vlue tht it converges to t tht point is the limit of f() s R from the right. For emple, the Mclurin series for tn given in (6) converges t both = nd =, since the hpotheses of the lternting series test (Theorem 9.6.) re stisfied t those points. Thus, the continuit of tn on the intervl [, ] implies tht t = the Mclurin series converges to nd t = it converges to lim tn = tn = π 4 lim + tn = tn ( ) = π 4 This shows tht the Mclurin series for tn ctull converges to tn on the closed intervl. Moreover, the convergence t = estblishes Formul ().

272 684 Chpter 9 / Infinite Series APPROXIMATING DEFINITE INTEGRALS USING TAYLOR SERIES Tlor series provide n lterntive to Simpson s rule nd other numericl methods for pproimting definite integrls. Emple 4 Approimte the integrl e d to three deciml-plce ccurc b epnding the integrnd in Mclurin series nd integrting term b term. Solution. We found in Emple () tht the Mclurin series for e is Therefore, e e d = = = + 4! 6 3! + 8 4! ] [ + 4! 6 3! + 8 4! d [ (!) 7 7(3!) + 9 9(4!) = 3 + 5! 7 3! + 9 4! ( ) k = (k + )k! k= Since this series clerl stisfies the hpotheses of the lternting series test (Theorem 9.6.), it follows from Theorem 9.6. tht if we pproimte the integrl b s n (the nth prtil sum of the series), then e d s n < [(n + ) + ](n + )! = (n + 3)(n + )! Thus, for three deciml-plce ccurc we must choose n such tht.5 = 5 4 (n + 3)(n + )! With the help of clculting utilit ou cn show tht the smllest vlue of n tht stisfies this condition is n = 5. Thus, the vlue of the integrl to three deciml-plce ccurc is ] Wht dvntges does the method of Emple 4 hve over Simpson s rule? Wht re its disdvntges? e d 3 + 5! 7 3! + 9 4! 5!.747 As check, clcultor with built-in numericl integrtion cpbilit produced the pproimtion.74684, which grees with our result when rounded to three deciml plces. FINDING TAYLOR SERIES BY MULTIPLICATION AND DIVISION The following emples illustrte some lgebric techniques tht re sometimes useful for finding Tlor series.

273 TECHNOLOGY MASTERY +... If ou hve CAS, use its cpbilit for multipling nd dividing polnomils to perform the computtions in Emples 9. Differentiting nd Integrting Power Series; Modeling with Tlor Series 685 Emple 5 Find the first three nonzero terms in the Mclurin series for the function f()= e tn. Solution. Using the series for e nd tn obtined in Emples nd 3 gives ) e tn = ( + )( Multipling, s shown in the mrgin, we obtin e tn = More terms in the series cn be obtined b including more terms in the fctors. Moreover, one cn prove tht series obtined b this method converges t ech point in the intersection of the intervls of convergence of the fctors (nd possibl on lrger intervl). Thus, we cn be certin tht the series we hve obtined converges for ll in the intervl (wh?). Emple 6 Find the first three nonzero terms in the Mclurin series for tn. Solution. Using the first three terms in the Mclurin series for sin nd cos, we cn epress tn s 3 tn = sin cos = 3! + 5 5!! + 4 4! Dividing, s shown in the mrgin, we obtin 5 nd 6. tn = Figure 9.. u L MODELING PHYSICAL LAWS WITH TAYLOR SERIES Tlor series provide n importnt w of modeling phsicl lws. To illustrte the ide we will consider the problem of modeling the period of simple pendulum (Figure 9..). As eplined in Chpter 7 Mking Connections Eercise 5, the period T of such pendulum is given b T = 4 L g π/ dφ (7) k sin φ where L = length of the supporting rod g = ccelertion due to grvit k = sin(θ /), where θ is the initil ngle of displcement from the verticl The integrl, which is clled complete elliptic integrl of the first kind, cnnot be epressed in terms of elementr functions nd is often pproimted b numericl methods. Unfortuntel, numericl vlues re so specific tht the often give little insight into generl phsicl principles. However, if we epnd the integrnd of (7) in series nd integrte term b term, then we cn generte n infinite series tht cn be used to construct vrious mthemticl models for the period T tht give deeper understnding of the behvior of the pendulum.

274 686 Chpter 9 / Infinite Series ACE STOCK LIMITED/Alm Understnding the motion of pendulum pled criticl role in the dvnce of ccurte timekeeping with the development of the pendulum clock in the seventeenth centur. To obtin series for the integrnd, we will substitute k sin φ for in the binomil series for / + tht we derived in Emple 4(b) of Section 9.9. If we do this, then we cn rewrite (7) s L π/ [ T = 4 + g k sin φ + 3! k4 sin 4 φ ] k 6 sin 6 φ + dφ (8) 3 3! If we integrte term b term, then we cn produce series tht converges to the period T. However, one of the most importnt cses of pendulum motion occurs when the initil displcement is smll, in which cse ll subsequent displcements re smll, nd we cn ssume tht k = sin(θ /). In this cse we epect the convergence of the series for T to be rpid, nd we cn pproimte the sum of the series b dropping ll but the constnt term in (8). This ields T = π which is clled the first-order model of T or the model for smll vibrtions. This model cn be improved on b using more terms in the series. For emple, if we use the first two terms in the series, we obtin the second-order model (verif). T = π L g L g ) ( + k 4 (9) () QUICK CHECK EXERCISES 9. (See pge 689 for nswers.). The Mclurin series for e obtined b substituting for in the series e k = k! k= 3. ( k= )( k ) k k! k + k= = ( + + )(! + + ) is e = k=.. [ ] d k+ k ( ) = d k k= = k= EXERCISE SET 9. C CAS. In ech prt, obtin the Mclurin series for the function b mking n pproprite substitution in the Mclurin series for /( ). Include the generl term in our nswer, nd stte the rdius of convergence of the series. () + (b) (c) (d). In ech prt, obtin the Mclurin series for the function b mking n pproprite substitution in the Mclurin series for ln( + ). Include the generl term in our nswer, nd = Suppose tht f() = 4 nd f ( ) k () = ( )k (k + )! k= () f () = (b) f() = + ( ) + ( ) + ( ) 3 + = + k= stte the rdius of convergence of the series. () ln( ) (b) ln( + ) (c) ln( + ) (d) ln( + ) 3. In ech prt, obtin the first four nonzero terms of the Mclurin series for the function b mking n pproprite substitution in one of the binomil series obtined in Emple 4 of Section 9.9. () ( + ) / (b) ( )

275 9. Differentiting nd Integrting Power Series; Modeling with Tlor Series () Use the Mclurin series for /( ) to find the Mclurin series for /( ), where =, nd stte the rdius of convergence of the series. (b) Use the binomil series for /( + ) obtined in Emple 4 of Section 9.9 to find the first four nonzero terms in the Mclurin series for /( + ), where =, nd stte the rdius of convergence of the series. 5 8 Find the first four nonzero terms of the Mclurin series for the function b mking n pproprite substitution in known Mclurin series nd performing n lgebric opertions tht re required. Stte the rdius of convergence of the series. 5. () sin (b) e (c) e (d) cos π 6. () cos (b) e (c) e (d) sin( ) 7. () () (b) sinh (c) ( ) 3/ (b) 3 cosh( ) (c) ( + ) 3 9 Find the first four nonzero terms of the Mclurin series for the function b using n pproprite trigonometric identit or propert of logrithms nd then substituting in known Mclurin series. 9. () sin (b) ln[( + 3 ] ( ). () cos (b) ln +. () Use known Mclurin series to find the Tlor series of / bout = b epressing this function s = ( ) (b) Find the intervl of convergence of the Tlor series.. Use the method of Eercise to find the Tlor series of / bout =, nd stte the intervl of convergence of the Tlor series. 3 4 Find the first four nonzero terms of the Mclurin series for the function b multipling the Mclurin series of the fctors. 3. () e sin (b) + ln( + ) 4. () e cos (b) ( + ) 4/3 ( + ) /3 5 6 Find the first four nonzero terms of the Mclurin series for the function b dividing pproprite Mclurin series. ( 5. () sec = ) (b) sin cos e 6. () tn ln( + ) (b) + 7. Use the Mclurin series for e nd e to derive the Mclurin series for sinh nd cosh. Include the generl terms in our nswers nd stte the rdius of convergence of ech series. 8. Use the Mclurin series for sinh nd cosh to obtin the first four nonzero terms in the Mclurin series for tnh. 9 Find the first five nonzero terms of the Mclurin series for the function b using prtil frctions nd known Mclurin series Confirm the derivtive formul b differentiting the pproprite Mclurin series term b term. d. () d [cos ] = sin (b) d [ln( + )]= d + d. () d [sinh ] =cosh (b) d d [tn ]= Confirm the integrtion formul b integrting the pproprite Mclurin series term b term. 3. () e d = e + C (b) sinh d= cosh + C 4. () sin d= cos + C (b) d = ln( + ) + C + 5. Consider the series k+ (k + )(k + ) k= Determine the intervls of convergence for this series nd for the series obtined b differentiting this series term b term. 6. Consider the series ( 3) k k k k= Determine the intervls of convergence for this series nd for the series obtined b integrting this series term b term. 7. () Use the Mclurin series for /( ) to find the Mclurin series for f() = (b) Use the Mclurin series obtined in prt () to find f (5) () nd f (6) (). (c) Wht cn ou s bout the vlue of f (n) ()? 8. Let f() = cos. Use the method of Eercise 7 to find f (99) (). 9 3 The limit of n indeterminte form s cn sometimes be found b epnding the functions involved in Tlor series bout = nd tking the limit of the series term b term. Use this method to find the limits in these eercises. sin tn 9. () lim (b) lim 3 3. () lim cos sin ln (b) lim + sin

276 688 Chpter 9 / Infinite Series 3 34 Use Mclurin series to pproimte the integrl to three deciml-plce ccurc sin( )d d 34. FOCUS ON CONCEPTS / / tn ( )d d () Find the Mclurin series for e 4. Wht is the rdius of convergence? (b) Eplin two different ws to use the Mclurin series for e 4 to find series for 3 e 4. Confirm tht both methods produce the sme series. 36. () Differentite the Mclurin series for /( ), nd use the result to show tht k k = for << ( ) k= (b) Integrte the Mclurin series for /( ), nd use the result to show tht k= k k = ln( ) for << (c) Use the result in prt (b) to show tht k+ k ( ) = ln( + ) for << k k= (d) Show tht the series in prt (c) converges if =. (e) Use the remrk following Emple 3 to show tht k+ k ( ) = ln( + ) for < k k= 37. Use the results in Eercise 36 to find the sum of the series. k () 3 = k k= (b) k(4 k ) = 4 + (4 ) + 3(4 3 ) + 4(4 4 ) + k= 38. Use the results in Eercise 36 to find the sum of ech series. () ( ) k+ k = (b) k= k= (e ) k = e (e ) (e )3 + + ke k e (e ) 3(e 3 ) 39. () Use the reltionship d = + sinh + C to find the first four nonzero terms in the Mclurin series for sinh. (b) Epress the series in sigm nottion. (c) Wht is the rdius of convergence? 4. () Use the reltionship d = sin + C to find the first four nonzero terms in the Mclurin series for sin. (b) Epress the series in sigm nottion. (c) Wht is the rdius of convergence? 4. We showed b Formul (9) of Section 8. tht if there re units of rdioctive crbon-4 present t time t =, then the number of units present t ers lter is (t) = e.t () Epress (t) s Mclurin series. (b) Use the first two terms in the series to show tht the number of units present fter er is pproimtel ( ). (c) Compre this to the vlue produced b the formul for (t). C 4. Suppose tht simple pendulum with length of L = meter is given n initil displcement of θ = 5 from the verticl. () Approimte the period T of the pendulum using Formul (9) for the first-order model of T. [Note: Tke g = 9.8 m/s.] (b) Approimte the period of the pendulum using Formul () for the second-order model. (c) Use the numericl integrtion cpbilit of CAS to pproimte the period of the pendulum from Formul (7), nd compre it to the vlues obtined in prts () nd (b). 43. Use the first three nonzero terms in Formul (8) nd the Wllis sine formul in the Endpper Integrl Tble (Formul ) to obtin model for the period of simple pendulum. 44. Recll tht the grvittionl force eerted b the Erth on n object is clled the object s weight (or more precisel, its Erth weight). If n object of mss m is on the surfce of the Erth (men se level), then the mgnitude of its weight is mg, where g is the ccelertion due to grvit t the Erth s surfce. A more generl formul for the mgnitude of the grvittionl force tht the Erth eerts on n object of mss m is F = mgr (R + h) where R is the rdius of the Erth nd h is the height of the object bove the Erth s surfce. () Use the binomil series for /( + ) obtined in Emple 4 of Section 9.9 to epress F s Mclurin series in powers of h/r. (b) Show tht if h =, then F = mg. (c) Show tht if h/r, then F mg (mgh/r). [Note: The quntit mgh/r cn be thought of s correction term for the weight tht tkes the object s height bove the Erth s surfce into ccount.] (d) If we ssume tht the Erth is sphere of rdius R = 4 mi t men se level, b pproimtel wht

277 Chpter 9 Review Eercises 689 percentge does person s weight chnge in going from men se level to the top of Mt. Everest (9,8 ft)? 45. () Show tht the Bessel function J () given b Formul (4) of Section 9.8 stisfies the differentil eqution + + =. (This is clled the Bessel eqution of order zero.) (b) Show tht the Bessel function J () given b Formul (5) of Section 9.8 stisfies the differentil eqution + + ( ) =. (This is clled the Bessel eqution of order one.) (c) Show tht J () = J (). 46. Prove: If the power series k= k k nd k= b k k hve the sme sum on n intervl ( r, r), then k = b k for ll vlues of k. 47. Writing Evlute the limit lim sin 3 in two ws: using L Hôpitl s rule nd b replcing sin b its Mclurin series. Discuss how the use of series cn give qulittive informtion bout how the vlue of n indeterminte limit is pproched. QUICK CHECK ANSWERS 9.. ( ) k k k!. ; ; ; ; ( ) k k 3. ; 3 ; () (b) 4; ; 4 ; ( )k ;4;( )k+ 8 k (k!) CHAPTER 9 REVIEW EXERCISES. Wht is the difference between n infinite sequence nd n infinite series?. Wht is ment b the sum of n infinite series? 3. () Wht is geometric series? Give some emples of convergent nd divergent geometric series. (b) Wht is p-series? Give some emples of convergent nd divergent p-series. 4. Stte conditions under which n lternting series is gurnteed to converge. 5. () Wht does it men to s tht n infinite series converges bsolutel? (b) Wht reltionship eists between convergence nd bsolute convergence of n infinite series? 6. Stte the Reminder Estimtion Theorem, nd describe some of its uses. 7. If power series in hs rdius of convergence R, wht cn ou s bout the set of -vlues t which the series converges? 8. () Write down the formul for the Mclurin series for f in sigm nottion. (b) Write down the formul for the Tlor series for f bout = in sigm nottion. 9. Are the following sttements true or flse? If true, stte theorem to justif our conclusion; if flse, then give counteremple. () If u k converges, then u k sk +. (b) If u k sk +, then u k converges. (c) If f(n) = n for n =,, 3,..., nd if n L s n +, then f() L s +. (d) If f(n) = n for n =,, 3,..., nd if f() L s +, then n L s n +. (e) If < n <, then { n } converges. (f ) If <u k <, then u k converges. (g) If u k nd v k converge, then (u k + v k ) diverges. (h) If u k nd v k diverge, then (u k v k ) converges. (i) If u k v k nd v k converges, then u k converges. (j) If u k v k nd u k diverges, then v k diverges. (k) If n infinite series converges, then it converges bsolutel. (l) If n infinite series diverges bsolutel, then it diverges.. Stte whether ech of the following is true or flse. Justif our nswers. () The function f() = /3 hs Mclurin series. (b) = (c) =. Find the generl term of the sequence, strting with n =, determine whether the sequence converges, nd if so find its limit. 3 (), 4 3, 5 4 3,... (b) 3, 5, 3 7, 4 9,.... Suppose tht the sequence { k } is defined recursivel b = c, k+ = k Assuming tht the sequence converges, find its limit if () c = (b) c = Show tht the sequence is eventull strictl monotone. { n } + () { (n ) 4} + (b) n= (n)!(n!) n= 4. () Give n emple of bounded sequence tht diverges. (b) Give n emple of monotonic sequence tht diverges. 5 Use n method to determine whether the series converge.

278 69 Chpter 9 / Infinite Series 5. () (b) 5 k 6. () 7. () 8. () 9. (). () k= ( ) k k + 4 k + k k 3 + k + ln k k= k= k= k= k= k k 9 k + k / + sin k (b) (b) (b) (b) (b) k= k= k= k= k= k= 5 k + ( ) k+ ( k + 3k (3 + k) /5 k 4/3 8k + 5k + cos(/k) k ( ) k+ k +. Find the ect error tht results when the sum of the geometric series k= ( /5) k is pproimted b the sum of the first terms in the series. n. Suppose tht u k = n. Find k= () u (b) lim k + u k (c) u k. 3. In ech prt, determine whether the series converges; if so, find its sum. ( 3 () ) (b) [ln(k + ) ln k] k 3 k k= k= (c) (d) [tn (k + ) tn k] k(k + ) k= k= 4. It cn be proved tht n lim n!=+ nd lim n + n + k= n n! n = e In ech prt, use these limits nd the root test to determine whether the series converges. k k k () (b) k! k! k= k= 5. Let, b, nd p be positive constnts. For which vlues of p does the series ( + bk) converge? p k= 6. Find the intervl of convergence of ( ) k (b > ) b k k= 7. () Show tht k k k!. (b) Use the comprison test to show tht ) k k k converges. (c) Use the root test to show tht the series converges. 8. Does the series converge? Justif 9 our nswer. k= 9. () Find the first five Mclurin polnomils of the function p() = (b) Mke generl sttement bout the Mclurin polnomils of polnomil of degree n. 3. Show tht the pproimtion sin 3 3! + 5 5! is ccurte to four deciml plces if π/4. 3. Use Mclurin series nd properties of lternting series to show tht ln( + ) /if<<. 3. Use Mclurin series to pproimte the integrl cos d to three deciml-plce ccurc. 33. In prts () (d), find the sum of the series b ssociting it with some Mclurin series. () + 4! + 8 3! + 6 4! + (b) π π3 3! + π5 5! π7 7! + (c) e! + e4 4! e6 6! + (ln 3) (ln 3)3 (d) ln 3 + +! 3! 34. In ech prt, write out the first four terms of the series, nd then find the rdius of convergence. 3 k () 4 7 (3k ) k k= (b) ( ) k 3 k 3 5 (k ) k+ k= 35. Use n ppropritetlor series for 3 to pproimte 3 8 to three deciml-plce ccurc, nd check our nswer b compring it to tht produced directl b our clculting utilit. 36. Differentite the Mclurin series for e nd use the result to show tht k + = e k! k= 37. Use the supplied Mclurin series for sin nd cos to find the first four nonzero terms of the Mclurin series for the given functions. sin = ( ) k k+ (k + )! k= cos = ( ) k k (k)! () sin cos k= (b) sin

279 Chpter 9 Mking Connections 69 CHAPTER 9 MAKING CONNECTIONS. As shown in the ccompning figure, suppose tht lines L nd L form n ngle θ, <θ<π/, t their point of intersection P. A point P is chosen tht is on L nd units from P. Strting from P zig-zg pth is constructed b successivel going bck nd forth between L nd L long perpendiculr from one line to the other. Find the following sums in terms of θ nd. () P P + P P + P P 3 + (b) P P + P P 3 + P 4 P 5 + (c) P P + P 3 P 4 + P 5 P 6 + L L P P 3 P 5 P P P 4 P 6 u P Figure E-. () Find A nd B such tht 6 k (3 k+ k+ )(3 k k ) = k A 3 k + k B k 3 k+ k+ (b) Use the result in prt () to find closed form for the nth prtil sum of the series 6 k (3 k+ k+ )(3 k k ) k= nd then find the sum of the series. Source: This eercise is dpted from problem tht ppered in the Fort-Fifth Annul Willim Lowell Putnm Competition. 3. Show tht the lternting p-series + p 3 p 4 p + +( )k+ k + p converges bsolutel if p>, converges conditionll if <p, nd diverges if p. 4. As illustrted in the ccompning figure, bug, strting t point A on 8 cm wire, wlks the length of the wire, stops nd wlks in the opposite direction for hlf the length of the wire, stops gin nd wlks in the opposite direction for onethird the length of the wire, stops gin nd wlks in the opposite direction for one-fourth the length of the wire, nd so forth until it stops for the th time. () Give upper nd lower bounds on the distnce between the bug nd point A when it finll stops. [Hint: As stted in Emple of Section 9.6, ssume tht the sum of the lternting hrmonic series is ln.] (b) Give upper nd lower bounds on the totl distnce tht the bug hs trveled when it finll stops. [Hint: Use inequlit () of Section 9.4.] A Figure E-4 8 cm 5. In Section 6.6 we defined the kinetic energ K of prticle with mss m nd velocit v to be K = mv [see Formul (7) of tht section]. In this formul the mss m is ssumed to be constnt, nd K is clled the Newtonin kinetic energ. However, in Albert Einstein s reltivit theor the mss m increses with the velocit nd the kinetic energ K is given b the formul [ ] K = m c (v /c) in which m is the mss of the prticle when its velocit is zero, nd c is the speed of light. This is clled the reltivistic kinetic energ. Use n pproprite binomil series to show tht if the velocit is smll compred to the speed of light (i.e., v/c ), then the Newtonin nd reltivistic kinetic energies re in close greement. 6. In Section 8.4 we studied the motion of flling object tht hs mss m nd is retrded b ir resistnce. We showed tht if the initil velocit is v nd the drg force F R is proportionl to the velocit, tht is, F R = cv, then the velocit of the object t time t is v(t) = e ct/m ( v + mg c ) mg c where g is the ccelertion due to grvit [see Formul (6) of Section 8.4]. () Use Mclurin series to show tht if ct/m, then the velocit cn be pproimted s ( cv ) v(t) v m + g t (b) Improve on the pproimtion in prt (). E XPANDING THE C ALCULUS H ORIZON To lern how ecologists use mthemticl models bsed on the process of itertion to stud the growth nd decline of niml popultions, see the module entitled Itertion nd Dnmicl Sstems t:

280 PARAMETRIC AND POLAR CURVES; CONIC SECTIONS Gilbert S. Grnt/Photo Reserchers, Inc. Mthemticl curves, such s the spirls in the center of sunflower, cn be described convenientl using ides developed in this chpter. In this chpter we will stud lterntive ws of epressing curves in the plne. We will begin b studing prmetric curves: curves described in terms of component functions. This stud will include methods for finding tngent lines to prmetric curves. We will then introduce polr coordinte sstems nd discuss methods for finding tngent lines to polr curves, rc length of polr curves, nd res enclosed b polr curves. Our ttention will then turn to review of the bsic properties of conic sections: prbols, ellipses, nd hperbols. Finll, we will consider conic sections in the contet of polr coordintes nd discuss some pplictions in stronom.. PARAMETRIC EQUATIONS; TANGENT LINES AND ARC LENGTH FOR PARAMETRIC CURVES Grphs of functions must pss the verticl line test, limittion tht ecludes curves with self-intersections or even such bsic curves s circles. In this section we will stud n lterntive method for describing curves lgebricll tht is not subject to the severe restriction of the verticl line test. We will then derive formuls required to find slopes, tngent lines, nd rc lengths of these prmetric curves. We will conclude with n investigtion of clssic prmetric curve known s the ccloid. (, ) C PARAMETRIC EQUATIONS Suppose tht prticle moves long curve C in the -plne in such w tht its - nd -coordintes, s functions of time, re = f(t), = g(t) We cll these the prmetric equtions of motion for the prticle nd refer to C s the trjector of the prticle or the grph of the equtions (Figure..). The vrible t is clled the prmeter for the equtions. A moving prticle with trjector C Figure.. Emple Sketch the trjector over the time intervl t of the prticle whose prmetric equtions of motion re = t 3 sin t, = 4 3 cos t () 69

281 . Prmetric Equtions; Tngent Lines nd Arc Length for Prmetric Curves 693 Solution. One w to sketch the trjector is to choose representtive succession of times, plot the (, ) coordintes of points on the trjector t those times, nd connect the points with smooth curve. The trjector in Figure.. ws obtined in this w from the dt in Tble.. in which the pproimte coordintes of the prticle re given t time increments of unit. Observe tht there is no t-is in the picture; the vlues of t pper onl s lbels on the plotted points, nd even these re usull omitted unless it is importnt to emphsize the loctions of the prticle t specific times. Tble.. t TECHNOLOGY MASTERY Red the documenttion for our grphing utilit to lern how to grph prmetric equtions, nd then generte the trjector in Emple. Eplore the behvior of the prticle beond time t =. 8 t = 3 t = 9 t = 6 t = 4 t = 4 t = 8 t = 5 t = t = 7 t = t = Figure Although prmetric equtions commonl rise in problems of motion with time s the prmeter, the rise in other contets s well. Thus, unless the problem dicttes tht the prmeter t in the equtions = f(t), = g(t) represents time, it should be viewed simpl s n independent vrible tht vries over some intervl of rel numbers. (In fct, there is no need to use the letter t for the prmeter; n letter not reserved for nother purpose cn be used.) If no restrictions on the prmeter re stted eplicitl or implied b the equtions, then it is understood tht it vries from to +. To indicte tht prmeter t is restricted to n intervl [,b], we will write = f(t), = g(t) ( t b) Emple Find the grph of the prmetric equtions = cos t, = sin t ( t π) () Figure..3 = cos t, = sin t ( t c) (, ) t (, ) Solution. One w to find the grph is to eliminte the prmeter t b noting tht + = sin t + cos t = Thus, the grph is contined in the unit circle + =. Geometricll, the prmeter t cn be interpreted s the ngle swept out b the rdil line from the origin to the point (, ) = (cos t,sin t) on the unit circle (Figure..3). As t increses from to π, the point trces the circle counterclockwise, strting t (, ) when t = nd completing one full revolution when t = π. One cn obtin different portions of the circle b vring the intervl over which the prmeter vries. For emple, = cos t, = sin t ( t π) (3) represents just the upper semicircle in Figure..3.

282 694 Chpter / Prmetric nd Polr Curves; Conic Sections t (, ) = cos( t), = sin( t) ( t c) Figure..4 (, ) ORIENTATION The direction in which the grph of pir of prmetric equtions is trced s the prmeter increses is clled the direction of incresing prmeter or sometimes the orienttion imposed on the curve b the equtions. Thus, we mke distinction between curve, which is set of points, nd prmetric curve, which is curve with n orienttion imposed on it b set of prmetric equtions. For emple, we sw in Emple tht the circle represented prmetricll b () is trced counterclockwise s t increses nd hence hs counterclockwise orienttion. As shown in Figures.. nd..3, the orienttion of prmetric curve cn be indicted b rrowheds. To obtin prmetric equtions for the unit circle with clockwise orienttion, we cn replce t b t in () nd use the identities cos( t) = cos t nd sin( t) = sin t. This ields = cos t, = sin t ( t π) Here, the circle is trced clockwise b point tht strts t (, ) when t = nd completes one full revolution when t = π (Figure..4). TECHNOLOGY MASTERY When prmetric equtions re grphed using clcultor, the orienttion cn often be determined b wtching the direction in which the grph is trced on the screen. However, mn computers grph so fst tht it is often hrd to discern the orienttion. See if ou cn use our grphing utilit to confirm tht (3) hs counterclockwise orienttion. 8 Emple 3 Grph the prmetric curve = t 3, = 6t 7 b eliminting the prmeter, nd indicte the orienttion on the grph. Figure..5 = t 3, = 6t 7 Solution. To eliminte the prmeter we will solve the first eqution for t s function of, nd then substitute this epression for t into the second eqution: t = ( ) ( + 3) = 6 ( ) ( + 3) 7 = 3 + Thus, the grph is line of slope 3 nd -intercept. To find the orienttion we must look to the originl equtions; the direction of incresing t cn be deduced b observing tht increses s t increses or b observing tht increses s t increses. Either piece of informtion tells us tht the line is trced left to right s shown in Figure..5. REMARK (, ) (, ) Not ll prmetric equtions produce curves with definite orienttions; if the equtions re bdl behved, then the point trcing the curve m lep round spordicll or move bck nd forth, filing to determine definite direction. For emple, if = sin t, = sin t then the point (, ) moves long the prbol =. However, the vlue of vries periodicll between nd, so the point (, ) moves periodicll bck nd forth long the prbol between the points (, ) nd (, ) (s shown in Figure..6). Lter in the tet we will discuss restrictions tht eliminte such errtic behvior, but for now we will just void such complictions. Figure..6 EXPRESSING ORDINARY FUNCTIONS PARAMETRICALLY An eqution = f() cn be epressed in prmetric form b introducing the prmeter t = ; this ields the prmetric equtions = t, = f(t)

283 . Prmetric Equtions; Tngent Lines nd Arc Length for Prmetric Curves 695 For emple, the portion of the curve = cos over the intervl [ π, π] cn be epressed prmetricll s = t, = cos t ( π t π) (Figure..7). t = c t = t = c 7 7 Figure..7 t = c t = c f 8 f Figure..8 If function f is one-to-one, then it hs n inverse function f. In this cse the eqution = f () is equivlent to = f(). We cn epress the grph of f in prmetric form b introducing the prmeter = t; this ields the prmetric equtions = f(t), = t For emple, Figure..8 shows the grph of f() = nd its inverse. The grph of f cn be repesented prmetricll s = t, = t 5 + t + nd the grph of f cn be represented prmetricll s = t 5 + t +, = t TANGENT LINES TO PARAMETRIC CURVES We will be concerned with curves tht re given b prmetric equtions = f(t), = g(t) in which f(t) nd g(t) hve continuous first derivtives with respect to t. It cn be proved tht if d/dt =, then is differentible function of, in which cse the chin rule implies tht d d = d /dt (4) d/dt This formul mkes it possible to find d/d directl from the prmetric equtions without eliminting the prmeter. Emple 4 Find the slope of the tngent line to the unit circle = cos t, = sin t ( t π) t the point where t = π/6 (Figure..9). Figure..9 c/6 Solution. From (4), the slope t generl point on the circle is d d = d /dt d/dt = cos t sin t Thus, the slope t t = π/6 is d d = cot π t=π/6 6 = 3 = cot t (5)

284 696 Chpter / Prmetric nd Polr Curves; Conic Sections Note tht Formul (5) mkes sense geometricll becuse the rdius from the origin to the point P(cos t,sin t) hs slope m = tn t. Thus the tngent line t P, being perpendiculr to the rdius, hs slope m = tn t = cot t (Figure..). It follows from Formul (4) tht the tngent line to prmetric curve will be horizontl t those points where d/dt = nd d/dt =, since d/d = t such points. Two different situtions occur when d/dt =. At points where d/dt = nd d/dt =, the right side of (4) hs nonzero numertor nd zero denomintor; we will gree tht the curve hs infinite slope nd verticl tngent line t such points. At points where d/dt nd d/dt re both zero, the right side of (4) becomes n indeterminte form; we cll such points singulr points. No generl sttement cn be mde bout the behvior of prmetric curves t singulr points; the must be nlzed cse b cse. Emple 5 In disstrous first flight, n eperimentl pper irplne follows the trjector of the prticle in Emple : = t 3 sin t, = 4 3 cos t (t ) O t P(cos t, sin t) but crshes into wll t time t = (Figure..). () At wht times ws the irplne fling horizontll? (b) At wht times ws it fling verticll? Rdius OP hs slope m = tn t. Figure.. Stnislovs Kirs/iStockphoto The complicted motion of pper irplne is best described mthemticll using prmetric equtions. Solution (). The irplne ws fling horizontll t those times when d/dt = nd d/dt =. From the given trjector we hve d dt = 3 sin t nd d dt = 3 cos t (6) Setting d/dt = ields the eqution 3 sin t =, or, more simpl, sin t =. This eqution hs four solutions in the time intervl t : t =, t = π, t = π, t = 3π Since d/dt = 3 cos t = for these vlues of t (verif), the irplne ws fling horizontll t times t =, t = π 3.4, t = π 6.8, nd t = 3π 9.4 which is consistent with Figure... Solution (b). The irplne ws fling verticll t those times when d/dt = nd d/dt =. Setting d/dt = in (6) ields the eqution 3 cos t = or cos t = 3 This eqution hs three solutions in the time intervl t (Figure..): t = cos 3, t = π cos 3, t = π + cos 3 t = t = 8 t = 3 t = 4 t = 8 t = 7 t = t = 9 t = t = 5 t = 6 3 cos 3 = cos t o Figure.. Figure..

285 Prmetric Equtions; Tngent Lines nd Arc Length for Prmetric Curves 697 Since d/dt = 3 sin t is not zero t these points (wh?), it follows tht the irplne ws fling verticll t times t = cos.3, t π.3 5.5, t π which gin is consistent with Figure... Emple 6 The curve represented b the prmetric equtions = t, = t 3 ( <t<+ ) is clled semicubicl prbol. The prmeter t cn be eliminted b cubing nd squring, from which it follows tht = 3. The grph of this eqution, shown in Figure..3, consists of two brnches: n upper brnch obtined b grphing = 3/ nd lower brnch obtined b grphing = 3/. The two brnches meet t the origin, which corresponds to t = in the prmetric equtions. This is singulr point becuse the derivtives d/dt = t nd d/dt = 3t re both zero there. Figure..3 WARNING = t, = t 3 ( < t < + ) Although it is true tht d d = d /dt d/dt ou cnnot conclude tht d /d is the quotient of d /dt nd d /dt. To illustrte tht this conclusion is erroneous, show tht for the prmetric curve in Emple 7, d d = d /dt t= d /dt t= Emple 7 Without eliminting the prmeter, find d/d nd d /d t (, ) nd (, ) on the semicubicl prbol given b the prmetric equtions in Emple 6. Solution. From (4) we hve nd from (4) pplied to = d/d we hve d d = d /dt d/dt = 3t = 3 t (t = ) (7) t d d = d d = d /dt d/dt = 3 / t = 3 4t Since the point (, ) on the curve corresponds to t = in the prmetric equtions, it follows from (7) nd (8) tht d d = 3 d nd t= d = 3 t= 4 Similrl, the point (, ) corresponds to t = in the prmetric equtions, so ppling (7) nd (8) gin ields d d = 3 d nd t= d = 3 t= 4 Note tht the vlues we obtined for the first nd second derivtives re consistent with the grph in Figure..3, since t (, ) on the upper brnch the tngent line hs positive slope nd the curve is concve up, nd t (, ) on the lower brnch the tngent line hs negtive slope nd the curve is concve down. Finll, observe tht we were ble to ppl Formuls (7) nd (8) for both t = nd t =, even though the points (, ) nd (, ) lie on different brnches. In contrst, hd we chosen to perform the sme computtions b eliminting the prmeter, we would hve hd to obtin seprte derivtive formuls for = 3/ nd = 3/. (8) ARC LENGTH OF PARAMETRIC CURVES The following result provides formul for finding the rc length of curve from prmetric equtions for the curve. Its derivtion is similr to tht of Formul (3) in Section 6.4 nd will be omitted.

286 698 Chpter / Prmetric nd Polr Curves; Conic Sections Formuls (4) nd (5) in Section 6.4 cn be viewed s specil cses of (9). For emple, Formul (4) in Section 6.4 cn be obtined from (9) b writing = f() prmetricll s = t, = f(t) nd Formul (5) in Section 6.4 cn be obtined b writing = g() prmetricll s = g(t), = t.. rc length formul for prmetric curves If no segment of the curve represented b the prmetric equtions = (t), = (t) ( t b) is trced more thn once s t increses from to b, nd if d/dt nd d/dt re continuous functions for t b, then the rc length L of the curve is given b b (d ) ( ) d L = + dt (9) dt dt Emple 8 equtions Use (9) to find the circumference of circle of rdius from the prmetric = cos t, = sin t ( t π) Solution. L = π (d ) + dt ( ) d dt = dt = π π ( sin t) + ( cos t) dt ] π dt = t = π THE CYCLOID (THE APPLE OF DISCORD) The results of this section cn be used to investigte curve known s ccloid. This curve, which is one of the most significnt in the histor of mthemtics, cn be generted b point on circle tht rolls long stright line (Figure..4). This curve hs fscinting histor, which we will discuss shortl; but first we will show how to obtin prmetric equtions for it. For this purpose, let us ssume tht the circle hs rdius nd rolls long the positive -is of rectngulr coordinte sstem. Let P(,) be the point on the circle tht trces the ccloid, nd ssume tht P is initill t the origin. We will tke s our prmeter the ngle θ tht is swept out b the rdil line to P s the circle rolls (Figure..4). It is stndrd here to regrd θ s positive, even though it is generted b clockwise rottion. The motion of P is combintion of the movement of the circle s center prllel to the -is nd the rottion of P bout the center. As the rdil line sweeps out n ngle θ, the point P trverses n rc of length θ, nd the circle moves distnce θ long the -is. Thus, s suggested b Figure..5, the center moves to the point (θ, ), nd the coordintes of P re = θ sin θ, = cos θ () These re the equtions of the ccloid in terms of the prmeter θ. One of the resons the ccloid is importnt in the histor of mthemtics is tht the stud of its properties helped to spur the development of erl versions of differentition nd integrtion. Work on the ccloid ws crried out b some of the most fmous nmes in seventeenth centur mthemtics, including Johnn nd Jkob Bernoulli, Descrtes, L Hôpitl, Newton, nd Leibniz. The curve ws nmed the ccloid b the Itlin mthemticin nd stronomer, Glileo, who spent over 4 ers investigting its properties. An erl problem of interest ws tht of constructing tngent lines to the ccloid. This problem ws first solved b Descrtes, nd then b Fermt, whom Descrtes hd chllenged with the question. A modern solution to this problem follows directl from the prmetric equtions () nd Formul (4). For emple, using Formul (4), it is strightforwrd to show tht the -intercepts of the ccloid re cusps nd tht there is horizontl tngent line to the ccloid hlfw between djcent -intercepts (Eercise 6).

287 . Prmetric Equtions; Tngent Lines nd Arc Length for Prmetric Curves 699 P(, ) u cos u P c o å Figure..4 A ccloid = cos u Figure..5 sin u = u sin u u P Figure..6 Q Another erl problem ws determining the rc length of n rch of the ccloid. This ws solved in 658 b the fmous British rchitect nd mthemticin, Sir Christopher Wren. He showed tht the rc length of one rch of the ccloid is ectl eight times the rdius of the generting circle. [For solution to this problem using Formul (9), see Eercise 7.] The ccloid is lso importnt historicll becuse it provides the solution to two fmous mthemticl problems the brchistochrone problem (from Greek words mening shortest time ) nd the tutochrone problem (from Greek words mening equl time ). The brchistochrone problem is to determine the shpe of wire long which bed might slide from point P to nother point Q, not directl below, in the shortest time. The tutochrone problem is to find the shpe of wire from P to Q such tht two beds strted t n points on the wire between P nd Q rech Q in the sme mount of time. The solution to both problems turns out to be n inverted ccloid (Figure..6). In June of 696, Johnn Bernoulli posed the brchistochrone problem in the form of chllenge to other mthemticins. At first, one might conjecture tht the wire should form stright line, since tht shpe results in the shortest distnce from P to Q. However, the inverted ccloid llows the bed to fll more rpidl t first, building up sufficient speed to rech Q in the shortest time, even though it trvels longer distnce. The problem ws solved b Newton, Leibniz, nd L Hôpitl, s well s b Johnn Bernoulli nd his older brother Jkob; it ws formulted nd solved incorrectl ers erlier b Glileo, who thought the nswer ws circulr rc. In fct, Johnn ws so impressed with his brother Jkob s solution tht he climed it to be his own. (This ws just one of mn disputes bout the ccloid tht eventull led to the curve being known s the pple of discord. ) One solution of the brchistochrone problem leds to the differentil eqution ( ( ) ) d + = () d where is positive constnt. We leve it s n eercise (Eercise 7) to show tht the ccloid provides solution to this differentil eqution. Newton s solution of the brchistochrone problem in his own hndwriting

288 7 Chpter / Prmetric nd Polr Curves; Conic Sections QUICK CHECK EXERCISES. (See pge 75 for nswers.). Find prmetric equtions for circle of rdius, centered t (3, 5).. The grph of the curve described b the prmetric equtions = 4t, = 3t + is stright line with slope nd -intercept. 3. Suppose tht prmetric curve C is given b the equtions = f(t), = g(t) for t. Find prmetric equtions for C tht reverse the direction the curve is trced s the prmeter increses from to. 4. To find d/d directl from the prmetric equtions = f(t), = g(t) we cn use the formul d/d =. 5. Let L be the length of the curve = ln t, = sin t ( t π) An integrl epression for L is. EXERCISE SET. Grphing Utilit C CAS. () B eliminting the prmeter, sketch the trjector over the time intervl t 5 of the prticle whose prmetric equtions of motion re = t, = t + (b) Indicte the direction of motion on our sketch. (c) Mke tble of - nd -coordintes of the prticle t times t =,,, 3, 4, 5. (d) Mrk the position of the prticle on the curve t the times in prt (c), nd lbel those positions with the vlues of t. Johnn (left) nd Jkob (right) Bernoulli Members of n mzing Swiss fmil tht included severl genertions of outstnding mthemticins nd scientists. Nikolus Bernoulli (63 78), druggist, fled from Antwerp to escpe religious persecution nd ultimtel settled in Bsel, Switzerlnd. There he hd three sons, Jkob I (lso clled Jcques or Jmes), Nikolus, nd Johnn I (lso clled Jen or John). The Romn numerls re used to distinguish fmil members with identicl nmes (see the fmil tree below). Following Newton nd Leibniz, the Bernoulli brothers, Jkob I nd Johnn I, re considered b some to be the two most importnt founders of clculus. Jkob I ws self-tught in mthemtics. His fther wnted him to stud for the ministr, but he turned to mthemtics nd in 686 becme professor t the Universit of Bsel. When he strted working in mthemtics, he knew nothing of Newton s nd Leibniz work. He eventull becme fmilir with Newton s results, but becuse so little of Leibniz work ws published, Jkob duplicted mn of Leibniz results. Jkob s ounger brother Johnn I ws urged to enter into business b his fther. Insted, he turned to medicine nd studied mthemtics under the guidnce of his older brother. He eventull becme mthemtics professor t Gröningen in Hollnd, nd then, when Jkob died in 75, Johnn succeeded him s mthemtics professor t Bsel. Throughout their lives, Jkob I nd Johnn I hd mutul pssion for criticizing ech other s work, which frequentl erupted into ugl confronttions. Leibniz tried to medite the disputes, but Jkob, who resented Leibniz superior intellect, ccused him of siding with Johnn, nd thus Leibniz becme entngled in the rguments. The brothers often worked on common problems tht the posed s chllenges to one nother. Johnn, interested in gining fme, often used unscrupulous mens to mke himself pper the origintor of his brother s results; Jkob occsionll retlited. Thus, it is often difficult to determine who deserves credit for mn results. However, both men mde mjor contributions to the development of clculus. In ddition to his work on clculus, Jkob helped estblish fundmentl principles in probbilit, including the Lw of Lrge Numbers, which is cornerstone of modern probbilit theor. Among the other members of the Bernoulli fmil, Dniel, son of Johnn I, is the most fmous. He ws professor of mthemtics t St. Petersburg Acdem in Russi nd subsequentl professor of ntom nd then phsics t Bsel. He did work in clculus nd probbilit, but is best known for his work in phsics. A bsic lw of fluid flow, clled Bernoulli s principle, is nmed in his honor. He won the nnul prize of the French Acdem times for work on vibrting strings, tides of the se, nd kinetic theor of gses. Johnn II succeeded his fther s professor of mthemtics t Bsel. His reserch ws on the theor of het nd sound. Nikolus I ws mthemticin nd lw scholr who worked on probbilit nd series. On the recommendtion of Leibniz, he ws ppointed professor of mthemtics t Pdu nd then went to Bsel s professor of logic nd then lw. Nikolus II ws professor of jurisprudence in Switzerlnd nd then professor of mthemtics t St. Petersburg Acdem. Johnn III ws professor of mthemtics nd stronom in Berlin nd Jkob II succeeded his uncle Dniel s professor of mthemtics t St. Petersburg Acdem in Russi. Trul n incredible fmil! Jkob I (654 75) (Jcques, Jmes) Nikolus Bernoulli (63 78) Nikolus Nikolus I ( ) Johnn I ( ) (Jen, John) Nikolus II Dniel (695 76) (7 78) Johnn II (7 79) Johnn III Jkob II (744 87) ( )

289 . Prmetric Equtions; Tngent Lines nd Arc Length for Prmetric Curves 7. () B eliminting the prmeter, sketch the trjector over the time intervl t of the prticle whose prmetric equtions of motion re = cos(πt), = sin(πt) (b) Indicte the direction of motion on our sketch. (c) Mke tble of - nd -coordintes of the prticle t times t =,.5,.5,.75,. (d) Mrk the position of the prticle on the curve t the times in prt (c), nd lbel those positions with the vlues of t. 3 Sketch the curve b eliminting the prmeter, nd indicte the direction of incresing t. 3. = 3t 4, = 6t + 4. = t 3, = 3t 7 ( t 3) 5. = cos t, = 5 sin t ( t π) 6. = t, = t = 3 + cos t, = + 4 sin t ( t π) 8. = sec t, = tn t (π t<3π/) 9. = cos t, = sin t ( π/ t π/). = 4t + 3, = 6t 9. = sin t, = 3 cos t ( t π/). = sec t, = tn t ( t<π/) 3 8 Find prmetric equtions for the curve, nd check our work b generting the curve with grphing utilit. 3. A circle of rdius 5, centered t the origin, oriented clockwise. 4. The portion of the circle + = tht lies in the third qudrnt, oriented counterclockwise. 5. A verticl line intersecting the -is t =, oriented upwrd. 6. The ellipse /4 + /9 =, oriented counterclockwise. 7. The portion of the prbol = joining (, ) nd (, ), oriented down to up. 8. The circle of rdius 4, centered t (, 3), oriented counterclockwise. 9. () Use grphing utilit to generte the trjector of prticle whose equtions of motion over the time intervl t 5 re = 6t t 3, = + t (b) Mke tble of - nd -coordintes of the prticle t times t =,,, 3, 4, 5. (c) At wht times is the prticle on the -is? (d) During wht time intervl is <5? (e) At wht time does the -coordinte of the prticle rech mimum?. () Use grphing utilit to generte the trjector of pper irplne whose equtions of motion for t re = t sin t, = 3 cos t (b) Assuming tht the plne flies in room in which the floor is t =, eplin wh the plne will not crsh into the floor. [For simplicit, ignore the phsicl size of the plne b treting it s prticle.] (c) How high must the ceiling be to ensure tht the plne does not touch or crsh into it? Grph the eqution using grphing utilit.. () = + + (b) = sin, π π. () = (b) = tn, π/ <<π/ FOCUS ON CONCEPTS 3. In ech prt, mtch the prmetric eqution with one of the curves lbeled (I) (VI), nd eplin our resoning. () = t, = sin 3t (b) = cos t, = 3 sin t (c) = t cos t, = t sin t (d) = 3t + t, = 3t 3 + t 3 (e) = t 3 + t, = t + t (f ) = cos t, = sin t I IV Figure E-3 II V 4. () Identif the orienttion of the curves in Eercise 3. (b) Eplin wh the prmetric curve = t, = t 4 ( t ) does not hve definite orienttion. 5. () Suppose tht the line segment from the point P(, ) to Q(, ) is represented prmetricll b = + ( )t, ( t ) = + ( )t nd tht R(,) is the point on the line segment corresponding to specified vlue of t (see the ccompning figure on the net pge). Show tht t = r/q, where r is the distnce from P to R nd q is the distnce from P to Q. (cont.) III VI

290 7 Chpter / Prmetric nd Polr Curves; Conic Sections (b) Wht vlue of t produces the midpoint between points P nd Q? (c) Wht vlue of t produces the point tht is three-fourths of the w from P to Q? t = P(, ) t R(, ) t = Q(, ) Figure E-5 6. Find prmetric equtions for the line segment joining P(, ) nd Q(3, ), nd use the result in Eercise 5 to find () the midpoint between P nd Q (b) the point tht is one-fourth of the w from P to Q (c) the point tht is three-fourths of the w from P to Q. 7. () Show tht the line segment joining the points (, ) nd (, ) cn be represented prmetricll s = + ( ) t t, t t (t t t ) = + ( ) t t t t (b) Which w is the line segment oriented? (c) Find prmetric equtions for the line segment trced from (3, ) to (, 4) s t vries from to, nd check our result with grphing utilit. 8. () B eliminting the prmeter, show tht if nd c re not both zero, then the grph of the prmetric equtions = t + b, = ct + d (t t t ) is line segment. (b) Sketch the prmetric curve = t, = t + ( t ) nd indicte its orienttion. (c) Wht cn ou s bout the line in prt () if or c (but not both) is zero? (d) Wht do the equtions represent if nd c re both zero? 9 3 Use grphing utilit nd prmetric equtions to displ the grphs of f nd f on the sme screen. 9. f() = 3 +., 3. f() = + +, f() = cos(cos.5), 3 3. f() = + sin, True Flse Determine whether the sttement is true or flse. Eplin our nswer. 33. The eqution = cn be described prmetricll b = sin t, = cos t. 34. The grph of the prmetric equtions = f(t), = t is the reflection of the grph of = f() bout the -is. 35. For the prmetric curve = (t), = 3t 4 t 3, the derivtive of with respect to is computed b d d = t 3 6t (t) 36. The curve represented b the prmetric equtions = t 3, = t + t 6 ( <t<+ ) is concve down for t<. 37. Prmetric curves cn be defined piecewise b using different formuls for different vlues of the prmeter. Sketch the curve tht is represented piecewise b the prmetric equtions { ( ) = t, = 4t t ( = t, = t t ) 38. Find prmetric equtions for the rectngle in the ccompning figure, ssuming tht the rectngle is trced counterclockwise s t vries from to, strting t (, ) when t =. [Hint: Represent the rectngle piecewise, letting t vr from to 4 for the first edge, from 4 to for the second edge, nd so forth.],,,, Figure E () Find prmetric equtions for the ellipse tht is centered t the origin nd hs intercepts (4, ), ( 4, ), (, 3), nd (, 3). (b) Find prmetric equtions for the ellipse tht results b trnslting the ellipse in prt () so tht its center is t (, ). (c) Confirm our results in prts () nd (b) using grphing utilit. 4. We will show lter in the tet tht if projectile is fired from ground level with n initil speed of v meters per second t n ngle α with the horizontl, nd if ir resistnce is neglected, then its position fter t seconds, reltive to the coordinte sstem in the ccompning figure on the net pge is = (v cos α)t, = (v sin α)t gt where g 9.8 m/s. () B eliminting the prmeter, show tht the trjector lies on the grph of qudrtic polnomil. (b) Use grphing utilit to sketch the trjector if α = 3 nd v = m/s. (c) Using the trjector in prt (b), how high does the shell rise? (cont.)

291 . Prmetric Equtions; Tngent Lines nd Arc Length for Prmetric Curves 73 (d) Using the trjector in prt (b), how fr does the shell trvel horizontll? FOCUS ON CONCEPTS Figure E-4 4. () Find the slope of the tngent line to the prmetric curve = t/, = t + tt = nd t t = without eliminting the prmeter. (b) Check our nswers in prt () b eliminting the prmeter nd differentiting n pproprite function of. 4. () Find the slope of the tngent line to the prmetric curve = 3 cos t, = 4 sin t t t = π/4 nd t t = 7π/4 without eliminting the prmeter. (b) Check our nswers in prt () b eliminting the prmeter nd differentiting n pproprite function of. 43. For the prmetric curve in Eercise 4, mke conjecture bout the sign of d /d t t = nd t t =, nd confirm our conjecture without eliminting the prmeter. 44. For the prmetric curve in Eercise 4, mke conjecture bout the sign of d /d t t = π/4 nd t t = 7π/4, nd confirm our conjecture without eliminting the prmeter Find ll vlues of t t which the prmetric curve hs () horizontl tngent line nd (b) verticl tngent line. 53. = sin t, = 4 cos t ( t π) 54. = t 3 5t + 4t + 7, = t + t In the mid-85s the French phsicist Jules Antoine Lissjous (8 88) becme interested in prmetric equtions of the form = sin t, = sin bt in the course of studing vibrtions tht combine two perpendiculr sinusoidl motions. If /b is rtionl number, then the combined effect of the oscilltions is periodic motion long pth clled Lissjous curve. () Use grphing utilit to generte the complete grph of the Lissjous curves corresponding to =, b = ; =, b = 3; = 3, b = 4; nd = 4, b = 5. (b) The Lissjous curve = sin t, = sin t ( t π) crosses itself t the origin (see Figure E-55). equtions for the two tngent lines t the origin. 56. The prolte ccloid = π cos t, = t π sin t ( π t π) Find crosses itself t point on the -is (see the ccompning figure). Find equtions for the two tngent lines t tht point Find d/d nd d /d t the given point without eliminting the prmeter. 45. = t, = t + 4; t = 46. = t +, = 3 t 3 t; t = 47. = sec t, = tn t; t = π/3 48. = sinh t, = cosh t; t = 49. = θ + cos θ, = + sin θ; θ = π/6 5. = cos φ, = 3 sin φ; φ = 5π/6 5. () Find the eqution of the tngent line to the curve = e t, = e t t t = without eliminting the prmeter. (b) Find the eqution of the tngent line in prt () b eliminting the prmeter. 5. () Find the eqution of the tngent line to the curve = t + 4, = 8t t + 4 t t = without eliminting the prmeter. (b) Find the eqution of the tngent line in prt () b eliminting the prmeter. Figure E-55 Figure E Show tht the curve = t, = t 3 4t intersects itself t the point (4, ), nd find equtions for the two tngent lines to the curve t the point of intersection. 58. Show tht the curve with prmetric equtions = t 3t + 5, = t 3 + t t + 9 intersects itself t the point (3, ), nd find equtions for the two tngent lines to the curve t the point of intersection. 59. () Use grphing utilit to generte the grph of the prmetric curve = cos 3 t, = sin 3 t ( t π) nd mke conjecture bout the vlues of t t which singulr points occur. (b) Confirm our conjecture in prt () b clculting pproprite derivtives. 6. Verif tht the ccloid described b Formul () hs cusps t its -intercepts nd horizontl tngent lines t midpoints between djcent -intercepts (see Figure..4).

292 74 Chpter / Prmetric nd Polr Curves; Conic Sections C 6. () Wht is the slope of the tngent line t time t to the trjector of the pper irplne in Emple 5? (b) Wht ws the irplne s pproimte ngle of inclintion when it crshed into the wll? 6. Suppose tht bee follows the trjector = t cos t, = sin t ( t ) () At wht times ws the bee fling horizontll? (b) At wht times ws the bee fling verticll? 63. Consider the fmil of curves described b the prmetric equtions = cos t + h, = b sin t + k ( t<π) where = nd b =. Describe the curves in this fmil if () h nd k re fied but nd b cn vr (b) nd b re fied but h nd k cn vr (c) = nd b =, but h nd k vr so tht h = k () Use grphing utilit to stud how the curves in the fmil = cos t, = cos t sin t ( π<t<π) chnge s vries from to 5. (b) Confirm our conclusion lgebricll. (c) Write brief prgrph tht describes our findings Find the ect rc length of the curve over the stted intervl. 65. = t, = 3 t 3 ( t ) 66. = t, = t 3/4 ( t 6) 67. = cos 3t, = sin 3t ( t π) 68. = sin t + cos t, = sin t cos t ( t π) 69. = e t (sin t + cos t), = e t (sin t cos t)( t ) ( ) 7. = sin t, = ln( t ) t 7. () Use Formul (9) to show tht the length L of one rch of ccloid is given b π L = ( cos θ)dθ (b) Use CAS to show tht L is eight times the rdius of the wheel tht genertes the ccloid (see the ccompning figure). L = 8 o Figure E-7 7. Use the prmetric equtions in Formul () to verif tht the ccloid provides one solution to the differentil eqution ( ( ) ) d + = d where is positive constnt. FOCUS ON CONCEPTS 73. The musement prk rides illustrted in the ccompning figure consist of two connected rotting rms of length n inner rm tht rottes counterclockwise t rdin per second nd n outer rm tht cn be progrmmed to rotte either clockwise t rdins per second (the Scrmbler ride) or counterclockwise t rdins per second (the Clpso ride). The center of the rider cge is t the end of the outer rm. () Show tht in the Scrmbler ride the center of the cge hs prmetric equtions = cos t + cos t, = sin t sin t (b) Find prmetric equtions for the center of the cge in the Clpso ride, nd use grphing utilit to confirm tht the center trces the curve shown in the ccompning figure. (c) Do ou think tht rider trvels the sme distnce in one revolution of the Scrmbler ride s in one revolution of the Clpso ride? Justif our conclusion. Scrmbler ride Figure E-73 Clpso ride 74. () If thred is unwound from fied circle while being held tut (i.e., tngent to the circle), then the end of the thred trces curve clled n involute of circle. Show tht if the circle is centered t the origin, hs rdius, nd the end of the thred is initill t the point (, ), then the involute cn be epressed prmetricll s = (cos θ + θ sin θ), = (sin θ θ cos θ) where θ is the ngle shown in prt () of the ccompning figure on the net pge. (b) Assuming tht the dog in prt (b) of the ccompning figure on the net pge unwinds its lesh while keeping it tut, for wht vlues of θ in the intervl θ π will the dog be wlking North? South? Est? West? (c) Use grphing utilit to generte the curve trced b the dog, nd show tht it is consistent with our nswer in prt (b).

293 . Polr Coordintes 75 () Figure E-74 u (, ) W N S (b) u 75 8 If f (t) nd g (t) re continuous functions, nd if no segment of the curve = f(t), = g(t) ( t b) is trced more thn once, then it cn be shown tht the re of the surfce generted b revolving this curve bout the -is is b ( ) d ( ) d S = π + dt dt dt nd the re of the surfce generted b revolving the curve bout the -is is S = b π ( ) d + dt ( ) d dt dt [The derivtions re similr to those used to obtin Formuls (4) nd (5) in Section 6.5.] Use the formuls bove in these eercises. E 75. Find the re of the surfce generted b revolving = t, = 3t ( t ) bout the -is. 76. Find the re of the surfce generted b revolving the curve = e t cos t, = e t sin t( t π/) bout the -is. 77. Find the re of the surfce generted b revolving the curve = cos t, = sin t( t π/) bout the -is. 78. Find the re of the surfce generted b revolving = 6t, = 4t ( t ) bout the -is. 79. B revolving the semicircle = r cos t, = r sin t ( t π) bout the -is, show tht the surfce re of sphere of rdius r is 4πr. 8. The equtions = φ sin φ, = cos φ ( φ π) represent one rch of ccloid. Show tht the surfce re generted b revolving this curve bout the -is is given b S = 64π /3. 8. Writing Consult pproprite reference works nd write n ess on Americn mthemticin Nthniel Bowditch ( ) nd his investigtion of Bowditch curves (better known s Lissjous curves; see Eercise 55). 8. Writing Wht re some of the dvntges of epressing curve prmetricll rther thn in the form = f()? QUICK CHECK ANSWERS.. = 3 + cos t, = 5 + sin t ( t π). π 5. ( /t) + cos tdt 3 4 ; = f( t), = g( t) 4. d/dt d/dt = g (t) f (t). POLAR COORDINATES Up to now we hve specified the loction of point in the plne b mens of coordintes reltive to two perpendiculr coordinte es. However, sometimes moving point hs specil ffinit for some fied point, such s plnet moving in n orbit under the centrl ttrction of the Sun. In such cses, the pth of the prticle is best described b its ngulr direction nd its distnce from the fied point. In this section we will discuss new kind of coordinte sstem tht is bsed on this ide. POLAR COORDINATE SYSTEMS A polr coordinte sstem in plne consists of fied point O, clled the pole (or origin), nd r emnting from the pole, clled the polr is. In such coordinte sstem

294 76 Chpter / Prmetric nd Polr Curves; Conic Sections O Pole u r Figure.. P(r, u) Polr is we cn ssocite with ech point P in the plne pir of polr coordintes (r, θ), where r is the distnce from P to the pole nd θ is n ngle from the polr is to the r OP (Figure..). The number r is clled the rdil coordinte of P nd the number θ the ngulr coordinte (or polr ngle)of P. In Figure.., the points (6,π/4), (5, π/3), (3, 5π/4), nd (4, π/6) re plotted in polr coordinte sstems. If P is the pole, then r =, but there is no clerl defined polr ngle. We will gree tht n rbitrr ngle cn be used in this cse; tht is, (,θ)re polr coordintes of the pole for ll choices of θ. (6, c/4) (5, c/3) 5c/4 c/4 c/3 (3, 5c/4) c/6 (4, c/6) Figure.. The polr coordintes of point re not unique. For emple, the polr coordintes ll represent the sme point (Figure..3). (, 7π/4), (, π/4), nd (, 5π/4) 7c/4 5c/4 c/4 Figure..3 (, 7c/4) (, c/4) (, 5c/4) Terminl side 5c/4 P(3, 5c/4) c/4 P( 3, c/4) Figure..4 Terminl side Polr is Polr is In generl, if point P hs polr coordintes (r, θ), then (r, θ + nπ) nd (r, θ nπ) re lso polr coordintes of P for n nonnegtive integer n. Thus, ever point hs infinitel mn pirs of polr coordintes. As defined bove, the rdil coordinte r of point P is nonnegtive, since it represents the distnce from P to the pole. However, it will be convenient to llow for negtive vlues of r s well. To motivte n pproprite definition, consider the point P with polr coordintes (3, 5π/4). As shown in Figure..4, we cn rech this point b rotting the polr is through n ngle of 5π/4 nd then moving 3 units from the pole long the terminl side of the ngle, or we cn rech the point P b rotting the polr is through n ngle of π/4 nd then moving 3 units from the pole long the etension of the terminl side. This suggests tht the point (3, 5π/4) might lso be denoted b ( 3,π/4), with the minus sign serving to indicte tht the point is on the etension of the ngle s terminl side rther thn on the terminl side itself. In generl, the terminl side of the ngle θ + π is the etension of the terminl side of θ, so we define negtive rdil coordintes b greeing tht re polr coordintes of the sme point. ( r, θ) nd (r, θ + π) RELATIONSHIP BETWEEN POLAR AND RECTANGULAR COORDINATES Frequentl, it will be useful to superimpose rectngulr -coordinte sstem on top of polr coordinte sstem, mking the positive -is coincide with the polr is. If this is done, then ever point P will hve both rectngulr coordintes (, ) nd polr coordintes

295 . Polr Coordintes 77 c/ P (, ) (r, u) (r, θ). As suggested b Figure..5, these coordintes re relted b the equtions = r cos θ, = r sin θ () r u = r cos u Figure..5 = r sin u These equtions re well suited for finding nd when r nd θ re known. However, to find r nd θ when nd re known, it is preferble to use the identities sin θ + cos θ = nd tn θ = sin θ/ cos θ to rewrite () s r = +, tn θ = () P r = 6 c/ c/3 Emple Find the rectngulr coordintes of the point P whose polr coordintes re (r, θ) = (6, π/3) (Figure..6). Solution. Substituting the polr coordintes r = 6 nd θ = π/3 in () ields = 6 cos π ( 3 = 6 ) = 3 ( ) = 6 sin π 3 3 = 6 = 3 3 Figure..6 Thus, the rectngulr coordintes of P re (, ) = ( 3, 3 3). c/ Emple Find polr coordintes of the point P whose rectngulr coordintes re (, 3 ) (Figure..7). Solution. We will find the polr coordintes (r, θ) of P tht stisf the conditions r> nd θ<π. From the first eqution in (), r = + = ( ) + ( 3 ) = 4 + = 6 so r = 4. From the second eqution in (), P(, 3) Figure..7 tn θ = = 3 = 3 From this nd the fct tht (, 3 ) lies in the third qudrnt, it follows tht the ngle stisfing the requirement θ<πis θ = 4π/3. Thus, (r, θ) = (4, 4π/3) re polr coordintes of P. All other polr coordintes of P re epressible in the form (4, 4π3 ) + nπ or ( 4, π ) 3 + nπ where n is n integer. GRAPHS IN POLAR COORDINATES We will now consider the problem of grphing equtions in r nd θ, where θ is ssumed to be mesured in rdins. Some emples of such equtions re r =, θ = π/4, r = θ, r = sin θ, r = cos θ In rectngulr coordinte sstem the grph of n eqution in nd consists of ll points whose coordintes (, ) stisf the eqution. However, in polr coordinte sstem, points hve infinitel mn different pirs of polr coordintes, so tht given point m hve some polr coordintes tht stisf n eqution nd others tht do not. Given n eqution

296 78 Chpter / Prmetric nd Polr Curves; Conic Sections in r nd θ, we define its grph in polr coordintes to consist of ll points with t lest one pir of coordintes (r, θ) tht stisf the eqution. Emple 3 Sketch the grphs of () r = (b) θ = π 4 in polr coordintes. Solution (). For ll vlues of θ, the point (,θ)is unit w from the pole. Since θ is rbitrr, the grph is the circle of rdius centered t the pole (Figure..8). Solution (b). For ll vlues of r, the point (r, π/4) lies on line tht mkes n ngle of π/4 with the polr is (Figure..8b). Positive vlues of r correspond to points on the line in the first qudrnt nd negtive vlues of r to points on the line in the third qudrnt. Thus, in bsence of n restriction on r, the grph is the entire line. Observe, however, tht hd we imposed the restriction r, the grph would hve been just the r in the first qudrnt. c/ c/ c/4 Figure..8 r = () u = c/4 (b) 9c/ 5c/ c/ Equtions r = f(θ) tht epress r s function of θ re especill importnt. One w to grph such n eqution is to choose some tpicl vlues of θ, clculte the corresponding vlues of r, nd then plot the resulting pirs (r, θ) in polr coordinte sstem. The net two emples illustrte this process. 3c c c 3c/ 4c Emple 4 Sketch the grph of r = θ(θ ) in polr coordintes b plotting points. Figure..9 7c/ r = u (u ) Solution. Observe tht s θ increses, so does r; thus, the grph is curve tht spirls out from the pole s θ increses. A resonbl ccurte sketch of the spirl cn be obtined b plotting the points tht correspond to vlues of θ tht re integer multiples of π/, keeping in mind tht the vlue of r is lws equl to the vlue of θ (Figure..9). Grph the spirl r = θ(θ ). Compre our grph to tht in Figure..9. Emple 5 points. Sketch the grph of the eqution r = sin θ in polr coordintes b plotting Solution. Tble.. shows the coordintes of points on the grph t increments of π/6. These points re plotted in Figure... Note, however, tht there re 3 points listed in the tble but onl 6 distinct plotted points. This is becuse the pirs from θ = π on ield

297 . Polr Coordintes 79 duplictes of the preceding points. For emple, ( /, 7π/6) nd (/,π/6) represent the sme point. Tble.. u (rdins) c e g i k m o r = sin u (r, u) (, ) π 3,, (, c) π 5π 6 7π π π 3 3, 5π, π 3 π 3 3, 3π 3 6 (, π) 3, 8, Figure.. r (, ), 6 r = sin u Rewriting this eqution s + = nd then completing the squre ields + ( ) = 4 which is circle of rdius centered t the point (, ) in the -plne. It is often useful to view the eqution r = f(θ)s n eqution in rectngulr coordintes (rther thn polr coordintes) nd grphed in rectngulr θr-coordinte sstem. For emple, Figure.. shows the grph of r = sin θ displed using rectngulr θr- coordintes. This grph cn ctull help to visulize how the polr grph in Figure.. is generted: c 6 i Figure.. r 6 Figure.. r = sin u c r = cos u i 3,, o o 4 u u Observe tht the points in Figure.. pper to lie on circle. We cn confirm tht this is so b epressing the polr eqution r = sin θ in terms of nd. To do this, we multipl the eqution through b r to obtin r = r sin θ which now llows us to ppl Formuls () nd () to rewrite the eqution s + = At θ = wehver =, which corresponds to the pole (, ) on the polr grph. As θ vries from to π/, the vlue of r increses from to, so the point (r, θ) moves long the circle from the pole to the high point t (,π/). As θ vries from π/ toπ, the vlue of r decreses from bck to, so the point (r, θ) moves long the circle from the high point bck to the pole. As θ vries from π to 3π/, the vlues of r re negtive, vring from to. Thus, the point (r, θ) moves long the circle from the pole to the high point t (,π/), which is the sme s the point (, 3π/). This duplictes the motion tht occurred for θ π/. As θ vries from 3π/toπ, the vlue of r vries from to. Thus, the point (r, θ) moves long the circle from the high point bck to the pole, duplicting the motion tht occurred for π/ θ π. Emple 6 Sketch the grph of r = cos θ in polr coordintes. Solution. Insted of plotting points, we will use the grph of r = cos θ in rectngulr coordintes (Figure..) to visulize how the polr grph of this eqution is generted. The nlsis nd the resulting polr grph re shown in Figure..3. This curve is clled four-petl rose.

298 7 Chpter / Prmetric nd Polr Curves; Conic Sections r vries from to s u vries from to c/4. r vries from to s u vries from c/4 to c/. r vries from to s u vries from c/ to 3c/4. r vries from to s u vries from 3c/4 to c. r vries from to s u vries from c to 5c/4. r vries from to s u vries from 5c/4 to 3c/. r vries from to s u vries from 3c/ to 7c/4. r vries from to s u vries from 7c/4 to c. Figure..3 SYMMETRY TESTS Observe tht the polr grph of r = cos θ in Figure..3 is smmetric bout the -is nd the -is. This smmetr could hve been predicted from the following theorem, which is suggested b Figure..4 (we omit the proof). The converse of ech prt of Theorem.. is flse. See Eercise theorem (Smmetr Tests) () (b) (c) A curve in polr coordintes is smmetric bout the -is if replcing θ b θ in its eqution produces n equivlent eqution (Figure..4). A curve in polr coordintes is smmetric bout the -is if replcing θ b π θ in its eqution produces n equivlent eqution (Figure..4b). A curve in polr coordintes is smmetric bout the origin if replcing θ b θ + π, or replcing r b r in its eqution produces n equivlent eqution (Figure..4c). c/ c/ c/ (r, u) (r, c u) (r, u) (r, u) (r, u) (r, u + c) or ( r, u) Figure..4 () (b) (c) Emple 7 Use Theorem.. to confirm tht the grph of r = cos θ in Figure..3 is smmetric bout the -is nd -is. A grph tht is smmetric bout both the -is nd the -is is lso smmetric bout the origin. Use Theorem..(c) to verif tht the curve in Emple 7 is smmetric bout the origin. Solution. To test for smmetr bout the -is, we replce θ b θ. This ields r = cos( θ) = cos θ Thus, replcing θ b θ does not lter the eqution. To test for smmetr bout the -is, we replce θ b π θ. This ields r = cos (π θ) = cos(π θ) = cos( θ) = cos θ Thus, replcing θ b π θ does not lter the eqution.

299 . Polr Coordintes 7 Emple 8 Sketch the grph of r = ( cos θ) in polr coordintes, ssuming to be positive constnt. Solution. Observe first tht replcing θ b θ does not lter the eqution, so we know in dvnce tht the grph is smmetric bout the polr is. Thus, if we grph the upper hlf of the curve, then we cn obtin the lower hlf b reflection bout the polr is. As in our previous emples, we will first grph the eqution in rectngulr θr-coordintes. This grph, which is shown in Figure..5, cn be obtined b rewriting the given eqution s r = cos θ, from which we see tht the grph in rectngulr θr-coordintes cn be obtined b first reflecting the grph of r = cos θ bout the -is to obtin the grph of r = cos θ, nd then trnslting tht grph up units to obtin the grph of r = cos θ. Now we cn see the following: As θ vries from to π/3, r increses from to /. As θ vries from π/3 toπ/, rincreses from / to. As θ vries from π/ toπ/3, rincreses from to 3/. As θ vries from π/3 toπ, r increses from 3/ to. This produces the polr curve shown in Figure..5b. The rest of the curve cn be obtined b continuing the preceding nlsis from π to π or, s noted bove, b reflecting the portion lred grphed bout the -is (Figure..5c). This hert-shped curve is clled crdioid (from the Greek word krdi mening hert ). 3 r 468 c o u (, c) 3, c 3 c,, c 3 r = ( cos u) () Figure..5 (b) (c) Emple 9 Sketch the grph of r = 4 cos θ in polr coordintes. Solution. This eqution does not epress r s function of θ,since solving for r in terms of θ ields two functions: r = cos θ nd r = cos θ Thus, to grph the eqution r = 4 cos θ we will hve to grph the two functions seprtel nd then combine those grphs. We will strt with the grph of r = cos θ. Observe first tht this eqution is not chnged if we replce θ b θ or if we replce θ b π θ. Thus, the grph is smmetric bout the -is nd the -is. This mens tht the entire grph cn be obtined b grphing the portion in the first qudrnt, reflecting tht portion bout the -is to obtin the portion in the second qudrnt, nd then reflecting those two portions bout the -is to obtin the portions in the third nd fourth qudrnts.

300 7 Chpter / Prmetric nd Polr Curves; Conic Sections To begin the nlsis, we will grph the eqution r = cos θ in rectngulr θrcoordintes (see Figure..6). Note tht there re gps in tht grph over the intervls π/4 <θ<3π/4 nd 5π/4 <θ<7π/4 becuse cos θ is negtive for those vlues of θ. From this grph we cn see the following: As θ vries from to π/4, rdecreses from to. As θ vries from π/4 toπ/, no points re generted on the polr grph. This produces the portion of the grph shown in Figure..6b. As noted bove, we cn complete the grph b reflection bout the -is followed b reflection bout the -is (Figure..6c). The resulting propeller-shped grph is clled lemniscte (from the Greek word lemniscos for looped ribbon resembling the number 8). We leve it for ou to verif tht the eqution r = cos θ hs the sme grph s r = cos θ,but trced in digonll opposite mnner. Thus, the grph of the eqution r = 4 cos θ consists of two identicl superimposed lemnisctes. r c/ u = c/4 c/ r = cos u u 3 9 f l o () (b) (c) Figure..6 c/ () u u = u FAMILIES OF LINES AND RAYS THROUGH THE POLE If θ is fied ngle, then for ll vlues of r the point (r, θ ) lies on the line tht mkes n ngle of θ = θ with the polr is; nd, conversel, ever point on this line hs pir of polr coordintes of the form (r, θ ). Thus, the eqution θ = θ represents the line tht psses through the pole nd mkes n ngle of θ with the polr is (Figure..7). If r is restricted to be nonnegtive, then the grph of the eqution θ = θ is the r tht emntes from the pole nd mkes n ngle of θ with the polr is (Figure..7b). Thus, s θ vries, the eqution θ = θ produces either fmil of lines through the pole or fmil of rs through the pole, depending on the restrictions on r. c/ FAMILIES OF CIRCLES We will consider three fmilies of circles in which is ssumed to be positive constnt: Figure..7 (b) u u = u (r ) r = r = cos θ r = sin θ (3 5) The eqution r = represents circle of rdius centered t the pole (Figure..8). Thus, s vries, this eqution produces fmil of circles centered t the pole. For fmilies (4) nd (5), recll from plne geometr tht tringle tht is inscribed in circle with dimeter of the circle for side must be right tringle. Thus, s indicted in Figures..8b nd..8c, the eqution r = cos θ represents circle of rdius, centered on the -is nd tngent to the -is t the origin; similrl, the eqution r = sin θ represents circle of rdius, centered on the -is nd tngent to the -is t the origin. Thus, s vries, Equtions (4) nd (5) produce the fmilies illustrted in Figures..8d nd..8e.

301 . Polr Coordintes 73 c/ c/ c/ c/ c/ P(, u) r u P(r, u) u r u P(r, u) r = r = cos u r = sin u r = cos u r = sin u () (b) (c) (d) (e) Figure..8 Observe tht replcing θ b θ does not chnge the eqution r = cos θ nd tht replcing θ b π θ does not chnge the eqution r = sin θ. This eplins wh the circles in Figure..8d re smmetric bout the -is nd those in Figure..8e re smmetric bout the -is. Wht do the grphs of one-petl roses look like? FAMILIES OF ROSE CURVES In polr coordintes, equtions of the form r = sin nθ r = cos nθ (6 7) in which > nd n is positive integer represent fmilies of flower-shped curves clled roses (Figure..9). The rose consists of n equll spced petls of rdius if n is odd nd n equll spced petls of rdius if n is even. It cn be shown tht rose with n even number of petls is trced out ectl once s θ vries over the intervl θ<π nd rose with n odd number of petls is trced out ectl once s θ vries over the intervl θ<π(eercise 78). A four-petl rose of rdius ws grphed in Emple 6. rose curves n = n = 3 n = 4 n = 5 n = 6 r = sin nu r = cos nu Figure..9 FAMILIES OF CARDIOIDS AND LIMAÇONS Equtions with n of the four forms r = ± b sin θ r = ± b cos θ (8 9) in which > nd b> represent polr curves clled limçons (from the Ltin word lim for snil-like creture tht is commonl clled slug ). There re four possible shpes for limçon tht re determined b the rtio /b (Figure..). If = b (the cse /b = ), then the limçon is clled crdioid becuse of its hert-shped ppernce, s noted in Emple 8.

302 74 Chpter / Prmetric nd Polr Curves; Conic Sections /b < /b = < /b < /b Figure.. Limçon with inner loop Crdioid Dimpled limçon Conve limçon Emple Figure.. shows the fmil of limçons r = + cos θ with the constnt vring from.5 to.5 in steps of.5. In keeping with Figure.., the limçons evolve from the loop tpe to the conve tpe. As increses from the strting vlue of.5, the loops get smller nd smller until the crdioid is reched t =. As increses further, the limçons evolve through the dimpled tpe into the conve tpe. =.5 =.5 =.75 = =.5 =.5 =.75 = =.5 =.5 Figure.. r = + cos u FAMILIES OF SPIRALS A spirl is curve tht coils round centrl point. Spirls generll hve left-hnd nd right-hnd versions tht coil in opposite directions, depending on the restrictions on the polr ngle nd the signs of constnts tht pper in their equtions. Some of the more common tpes of spirls re shown in Figure.. for nonnegtive vlues of θ,, nd b. c/ c/ c/ c/ c/ Archimeden spirl r = u Prbolic spirl r = u Logrithmic spirl r = e bu Lituus spirl r = / u Hperbolic spirl r = /u Figure.. SPIRALS IN NATURE Spirls of mn kinds occur in nture. For emple, the shell of the chmbered nutilus (below) forms logrithmic spirl, nd coiled silor s rope forms n Archimeden spirl. Spirls lso occur in flowers, the tusks of certin nimls, nd in the shpes of glies.

303 . Polr Coordintes 75 Michel Siu/iStockphoto The shell of the chmbered nutilus revels logrithmic spirl. The niml lives in the outermost chmber. Michel Thompson/iStockphoto A silor s coiled rope forms n Archimeden spirl. Courtes NASA & The Hubble Heritge Tem A spirl gl. GENERATING POLAR CURVES WITH GRAPHING UTILITIES For polr curves tht re too complicted for hnd computtion, grphing utilities cn be used. Although mn grphing utilities re cpble of grphing polr curves directl, some re not. However, if grphing utilit is cpble of grphing prmetric equtions, then it cn be used to grph polr curve r = f(θ) b converting this eqution to prmetric form. This cn be done b substituting f(θ) for r in (). This ields = f(θ)cos θ, = f(θ)sin θ () which is pir of prmetric equtions for the polr curve in terms of the prmeter θ. Emple Epress the polr eqution r = + cos 5θ prmetricll, nd generte the polr grph from the prmetric equtions using grphing utilit. Solution. Substituting the given epression for r in = r cos θ nd = r sin θ ields the prmetric equtions [ = + cos 5θ ] [ cos θ, = + cos 5θ ] sin θ TECHNOLOGY MASTERY Use grphing utilit to duplicte the curve in Figure..3. If our grphing utilit requires tht t be used s the prmeter, then ou will hve to replce θ b t in () to generte the grph. Net, we need to find n intervl over which to vr θ to produce the entire grph. To find such n intervl, we will look for the smllest number of complete revolutions tht must occur until the vlue of r begins to repet. Algebricll, this mounts to finding the smllest positive integer n such tht ( ) 5(θ + nπ) + cos = + cos 5θ or ( ) 5θ cos + 5nπ = cos 5θ

304 76 Chpter / Prmetric nd Polr Curves; Conic Sections c/ For this equlit to hold, the quntit 5nπ must be n even multiple of π; the smllest n for which this occurs is n =. Thus, the entire grph will be trced in two revolutions, which mens it cn be generted from the prmetric equtions = [ + cos 5θ ] cos θ, = This ields the grph in Figure..3. [ + cos 5θ ] sin θ ( θ 4π) r = + cos 5u Figure..3 QUICK CHECK EXERCISES. (See pge 79 for nswers.). () Rectngulr coordintes of point (,) m be recovered from its polr coordintes (r, θ) b mens of the equtions = nd =. (b) Polr coordintes (r, θ) m be recovered from rectngulr coordintes (,) b mens of the equtions r = nd tn θ =.. Find the rectngulr coordintes of the points whose polr coordintes re given. () (4,π/3) (b) (, π/6) (c) (6, π/3) (d) (4, 5π/4) 3. In ech prt, find polr coordintes stisfing the stted conditions for the point whose rectngulr coordintes re (, 3 ). () r nd θ<π (b) r nd θ<π 4. In ech prt, stte the nme tht describes the polr curve most precisel: rose, line, circle, limçon, crdioid, spirl, lemniscte, or none of these. () r = θ (b) r = + sin θ (c) r = sin θ (d) r = cos θ (e) r = csc θ (f ) r = + cos θ (g) r = sin θ EXERCISE SET. Grphing Utilit Plot the points in polr coordintes.. () (3,π/4) (b) (5, π/3) (c) (,π/) (d) (4, 7π/6) (e) ( 6, π) (f ) (, 9π/4). () (, π/3) (b) (3/, 7π/4) (c) ( 3, 3π/) (d) ( 5, π/6) (e) (, 4π/3) (f ) (,π) 3 4 Find the rectngulr coordintes of the points whose polr coordintes re given. 3. () (6,π/6) (b) (7, π/3) (c) ( 6, 5π/6) (d) (, π) (e) (7, 7π/6) (f ) ( 5, ) 4. () (,π/4) (b) (6, π/4) (c) (4, 9π/4) (d) (3, ) (e) ( 4, 3π/) (f ) (, 3π) 5. In ech prt, point is given in rectngulr coordintes. Find two pirs of polr coordintes for the point, one pir stisfing r nd θ<π, nd the second pir stisfing r nd π <θ. () ( 5, ) (b) ( 3, ) (c) (, ) (d) ( 8, 8) (e) ( 3, 3 3 ) (f ) (, ) 6. In ech prt, find polr coordintes stisfing the stted conditions for the point whose rectngulr coordintes re ( 3, ). () r nd θ<π (b) r nd θ<π (c) r nd π <θ (d) r nd π <θ π 7 8 Use clculting utilit, where needed, to pproimte the polr coordintes of the points whose rectngulr coordintes re given. 7. () (3, 4) (b) (6, 8) (c) (, tn ) 8. () ( 3, 4) (b) ( 3,.7) (c) ( ), sin 9 Identif the curve b trnsforming the given polr eqution to rectngulr coordintes. 9. () r = (b) r sin θ = 4 (c) r = 3 cos θ (d) r = 6 3 cos θ + sin θ. () r = 5 sec θ (b) r = sin θ (c) r = 4 cos θ + 4 sin θ (d) r = sec θ tn θ Epress the given equtions in polr coordintes.. () = 3 (b) + = 7 (c) = (d) 9 = 4

305 . Polr Coordintes 77. () = 3 (b) + = 5 (c) = (d) ( + ) = FOCUS ON CONCEPTS 3 6 A grph is given in rectngulr θr-coordinte sstem. Sketch the corresponding grph in polr coordintes. 3. r 4. r c o r u u 4 r 6 c 7 Find n eqution for the given polr grph. [Note: Numeric lbels on these grphs represent distnces to the origin.] 7. () (b) (c) 5 6 c 6 Circle Circle Crdioid 8. () (b) (c) 3 Limçon Circle Three-petl rose 9. () 3 (b) 5 (c) Four-petl rose Limçon Lemniscte. () (b) (c) i o u u 3 7. r = 3( + sin θ) 8. r = 5 5 sin θ 9. r = 4 4 cos θ 3. r = + sin θ 3. r = cos θ 3. r = cos θ 33. r = 3 sin θ 34. r = cos θ 35. r 5 = 3 sin θ 36. r = 5 cos θ 37. r = 3 4 sin θ 38. r = cos θ 39. r = 6 sin θ 4. r = 4θ (θ ) 4. r = 4θ (θ ) 4. r = 4θ 43. r = cos θ 44. r = 3 sin θ 45. r = 9 sin 4θ 46. r = cos 3θ 47 5 True Flse Determine whether the sttement is true or flse. Eplin our nswer. 47. The polr coordinte pirs (,π/3) nd (, π/3) describe the sme point. 48. If the grph of r = f(θ) drwn in rectngulr θrcoordintes is smmetric bout the r-is, then the grph of r = f(θ) drwn in polr coordintes is smmetric bout the -is. 49. The portion of the polr grph of r = sin θ for vlues of θ between π/ nd π is contined in the second qudrnt. 5. The grph of dimpled limçon psses through the polr origin Determine shortest prmeter intervl on which complete grph of the polr eqution cn be generted, nd then use grphing utilit to generte the polr grph. 5. r = cos θ 53. r = sin θ 4 5. r = sin θ 54. r =.5 + cos θ r = cos θ The ccompning figure shows the grph of the butterfl curve r = e cos θ cos 4θ + sin 3 θ 4 Determine shortest prmeter intervl on which the complete butterfl cn be generted, nd then check our nswer using grphing utilit. c/ 6 Crdioid Five-petl rose Circle 46 Sketch the curve in polr coordintes.. θ = π 3. θ = 3π 4 3. r = 3 4. r = 4 cos θ 5. r = 6 sin θ 6. r = cos θ Figure E-56

306 78 Chpter / Prmetric nd Polr Curves; Conic Sections 57. The ccompning figure shows the Archimeden spirl r = θ/ produced with grphing clcultor. () Wht intervl of vlues for θ do ou think ws used to generte the grph? (b) Duplicte the grph with our own grphing utilit. 6 6 A polr grph of r = f(θ) is given over the stted intervl. Sketch the grph of ( () r = f( θ) (b) r = f θ π ) ( (c) r = f θ + π ) (d) r = f(θ). 6. θ π/ 6. π/ θ π c/ (, c/4) (, 3c/4) c/ [ 9, 9] [ 6, 6] Scl =, Scl = Figure E-57 Figure E-6 Figure E Find equtions for the two fmilies of circles in the ccompning figure. c/ I Figure E () Show tht if vries, then the polr eqution r = sec θ ( π/ <θ<π/) describes fmil of lines perpendiculr to the polr is. (b) Show tht if b vries, then the polr eqution r = b csc θ ( <θ<π) describes fmil of lines prllel to the polr is. FOCUS ON CONCEPTS II c/ 6. The ccompning figure shows grphs of the Archimeden spirl r = θ nd the prbolic spirl r = θ. Which is which? Eplin our resoning. c/ c/ Use the polr grph from the indicted eercise to sketch the grph of () r = f(θ) + (b) r = f(θ). 63. Eercise Eercise Show tht if the polr grph of r = f(θ) is rotted counterclockwise round the origin through n ngle α, then r = f(θ α) is n eqution for the rotted curve. [Hint: If (r,θ ) is n point on the originl grph, then (r,θ + α) is point on the rotted grph.] 66. Use the result in Eercise 65 to find n eqution for the lemniscte tht results when the lemniscte in Emple 9 is rotted counterclockwise through n ngle of π/. 67. Use the result in Eercise 65 to find n eqution for the crdioid r = + cos θ fter it hs been rotted through the given ngle, nd check our nswer with grphing utilit. () π (b) π (c) π (d) 5π () Show tht if A nd B re not both zero, then the grph of the polr eqution r = A sin θ + B cos θ is circle. Find its rdius. (b) Derive Formuls (4) nd (5) from the formul given in prt (). 69. Find the highest point on the crdioid r = + cos θ. 7. Find the leftmost point on the upper hlf of the crdioid r = + cos θ. 7. Show tht in polr coordinte sstem the distnce d between the points (r,θ ) nd (r,θ ) is d = r + r r r cos(θ θ ) I Figure E-6 II 7 74 Use the formul obtined in Eercise 7 to find the distnce between the two points indicted in polr coordintes. 7. (3,π/6) nd (,π/3)

307 .3 Tngent Lines, Arc Length, nd Are for Polr Curves Successive tips of the four-petl rose r = cos θ. Check our nswer using geometr. 74. Successive tips of the three-petl rose r = sin 3θ. Check our nswer using trigonometr. 75. In the lte seventeenth centur the Itlin stronomer Giovnni Domenico Cssini (65 7) introduced the fmil of curves ( + + ) b 4 4 = ( >,b >) in his studies of the reltive motions of the Erth nd the Sun. These curves, which re clled Cssini ovls, hve one of the three bsic shpes shown in the ccompning figure. () Show tht if = b, then the polr eqution of the Cssini ovl is r = cos θ, which is lemniscte. (b) Use the formul in Eercise 7 to show tht the lemniscte in prt () is the curve trced b point tht moves in such w tht the product of its distnces from the polr points (, ) nd (, π) is. Figure E-75 = b < b > b 76. Show tht the hperbolic spirl r = /θ (θ>) hs horizontl smptote t = b showing tht nd + s θ +. Confirm this result b generting the spirl with grphing utilit. 77. Show tht the spirl r = /θ does not hve n horizontl smptotes. 78. Prove tht rose with n even number of petls is trced out ectl once s θ vries over the intervl θ<πnd rose with n odd number of petls is trced out ectl once s θ vries over the intervl θ<π. 79. () Use grphing utilit to confirm tht the grph of r = sin(θ/)( θ 4π) is smmetric bout the -is. (b) Show tht replcing θ b θ in the polr eqution r = sin(θ/) does not produce n equivlent eqution. Wh does this not contrdict the smmetr demonstrted in prt ()? 8. Writing Use grphing utilit to investigte how the fmil of polr curves r = + cos nθ is ffected b chnging the vlues of nd n, where is positive rel number nd n is positive integer. Write brief prgrph to eplin our conclusions. 8. Writing Wh do ou think the djective polr ws chosen in the nme polr coordintes? Verticl nd horizontl smptotes of polr curves cn sometimes be detected b investigting the behvior of = r cos θ nd = r sin θ s θ vries. This ide is used in these eercises. QUICK CHECK ANSWERS.. () r cos θ; r sin θ (b) + ; /. () (, 3) (b) ( 3, ) (c) ( 3, 3 3) (d) (, ) 3. () (,π/3) (b) (, 4π/3) 4. () spirl (b) limçon (c) rose (d) none of these (e) line (f ) crdioid (g) circle.3 TANGENT LINES, ARC LENGTH, AND AREA FOR POLAR CURVES In this section we will derive the formuls required to find slopes, tngent lines, nd rc lengths of polr curves. We will then show how to find res of regions tht re bounded b polr curves. TANGENT LINES TO POLAR CURVES Our first objective in this section is to find method for obtining slopes of tngent lines to polr curves of the form r = f(θ) in which r is differentible function of θ. We showed in the lst section tht curve of this form cn be epressed prmetricll in terms of the prmeter θ b substituting f(θ) for r in the equtions = r cos θ nd = r sin θ. This ields = f(θ)cos θ, = f(θ)sin θ

308 7 Chpter / Prmetric nd Polr Curves; Conic Sections from which we obtin d dθ = f(θ)sin θ + f (θ) cos θ = r sin θ + dr dθ cos θ d dθ = f(θ)cos θ + f (θ) sin θ = r cos θ + dr dθ sin θ Thus, if d/dθ nd d/dθ re continuous nd if d/dθ =, then is differentible function of, nd Formul (4) in Section. with θ in plce of t ields () dr d d = d /dθ r cos θ + sin θ d/dθ = dθ r sin θ + cos θ dr dθ () Emple where θ = π/4. Find the slope of the tngent line to the circle r = 4 cos θ t the point c/ c/4 Tngent 4 Solution. From () with r = 4 cos θ, so tht dr/dθ = 4 sin θ, we obtin d d = 4 cos θ 4 sin θ 8 sin θ cos θ Using the double-ngle formuls for sine nd cosine, d d = cos θ sin θ = cos θ sin θ sin θ cos θ = cot θ Thus, t the point where θ = π/4 the slope of the tngent line is m = d d = cot π θ=π/4 = r = 4 cos u which implies tht the circle hs horizontl tngent line t the point where θ = π/4 Figure.3. (Figure.3.). Emple Find the points on the crdioid r = cos θ t which there is horizontl tngent line, verticl tngent line, or singulr point. Solution. A horizontl tngent line will occur where d/dθ = nd d/dθ =, verticl tngent line where d/dθ = nd d/dθ =, nd singulr point where d/dθ = nd d/dθ =. We could find these derivtives from the formuls in (). However, n lterntive pproch is to go bck to bsic principles nd epress the crdioid prmetricll b substituting r = cos θ in the conversion formuls = r cos θ nd = r sin θ. This ields = ( cos θ)cos θ, = ( cos θ)sin θ ( θ π) Differentiting these equtions with respect to θ nd then simplifing ields (verif) d dθ = sin θ( cos θ ), d dθ = ( cos θ)( + cos θ) Thus, d/dθ = if sin θ = or cos θ =, nd d /dθ = if cos θ = orcosθ =. We leve it for ou to solve these equtions nd show tht the solutions of d/dθ = on the intervl θ π re d dθ = : θ =, π 3, π, 5π 3, π

309 .3 Tngent Lines, Arc Length, nd Are for Polr Curves 7 c/ nd the solutions of d/dθ = on the intervl θ π re (, c) 3 c, 3 3 4c, 3, c 3 5c, 3 d dθ = : θ =, π 3, 4π 3, π Thus, horizontl tngent lines occur t θ = π/3 nd θ = 4π/3; verticl tngent lines occur t θ = π/3, π,nd 5π/3; nd singulr points occur t θ = nd θ = π (Figure.3.). Note, however, tht r = t both singulr points, so there is rell onl one singulr point on the crdioid the pole. r = cos u Figure.3. c/ r = f(u) TANGENT LINES TO POLAR CURVES AT THE ORIGIN Formul () revels some useful informtion bout the behvior of polr curve r = f(θ) tht psses through the origin. If we ssume tht r = nd dr/dθ = when θ = θ, then it follows from Formul () tht the slope of the tngent line to the curve t θ = θ is dr d + sin θ d = dθ dr + cos θ dθ = sin θ cos θ = tn θ (Figure.3.3). However, tn θ is lso the slope of the line θ = θ, so we cn conclude tht this line is tngent to the curve t the origin. Thus, we hve estblished the following result. u = u Slope = tn u u Figure theorem If the polr curve r = f(θ) psses through the origin t θ = θ, nd if dr/dθ = t θ = θ, then the line θ = θ is tngent to the curve t the origin. This theorem tells us tht equtions of the tngent lines t the origin to the curve r = f(θ) cn be obtined b solving the eqution f(θ) =. It is importnt to keep in mind, however, tht r = f(θ) m be zero for more thn one vlue of θ, so there m be more thn one tngent line t the origin. This is illustrted in the net emple. c/ u = c/3 u = c/3 Emple 3 The three-petl rose r = sin 3θ in Figure.3.4 hs three tngent lines t the origin, which cn be found b solving the eqution sin 3θ = It ws shown in Eercise 78 of Section. tht the complete rose is trced once s θ vries over the intervl θ<π,so we need onl look for solutions in this intervl. We leve it for ou to confirm tht these solutions re Figure.3.4 r = sin 3u θ =, θ = π 3, nd θ = π 3 Since dr/dθ = 3 cos 3θ = for these vlues of θ, these three lines re tngent to the rose t the origin, which is consistent with the figure. ARC LENGTH OF A POLAR CURVE A formul for the rc length of polr curve r = f(θ) cn be derived b epressing the curve in prmetric form nd ppling Formul (9) of Section. for the rc length of prmetric curve. We leve it s n eercise to show the following.

310 7 Chpter / Prmetric nd Polr Curves; Conic Sections.3. rc length formul for polr curves If no segment of the polr curve r = f(θ) is trced more thn once s θ increses from α to β, nd if dr/dθ is continuous for α θ β, then the rc length L from θ = α to θ = β is L = β α [f(θ)] +[f (θ)] dθ = β α r + ( ) dr dθ (3) dθ r = e u c/ Emple 4 Find the rc length of the spirl r = e θ in Figure.3.5 between θ = nd θ = π. (c, e c ) Figure.3.5 (, ) Solution. L = = β α π r + ( ) dr dθ = dθ π (eθ ) + (e θ ) dθ e θ dθ = ] π e θ = (e π ) 3.3 Emple 5 Find the totl rc length of the crdioid r = + cos θ. c/ r = + cos u Solution. The crdioid is trced out once s θ vries from θ = toθ = π. Thus, β ( ) dr π L = r + dθ = ( + cos θ) + ( sin θ) α dθ dθ = π + cos θdθ = = π π cos θdθ Identit (45) ofappendi B cos θ dθ Since cos θ chnges sign t π, we must split the lst integrl into the sum of two integrls: the integrl from to π plus the integrl from π to π. However, the integrl from π to π is equl to the integrl from to π, since the crdioid is smmetric bout the polr is (Figure.3.6). Thus, Figure.3.6 L = π π ] π cos θ dθ = 4 cos θdθ= 8 sin θ = 8 AREA IN POLAR COORDINATES We begin our investigtion of re in polr coordintes with simple cse. R r = f(u).3.3 re problem in polr coordintes Suppose tht α nd β re ngles tht stisf the condition α<β α + π u = b Figure.3.7 u = nd suppose tht f(θ) is continuous nd nonnegtive for α θ β. Find the re of the region R enclosed b the polr curve r = f(θ) nd the rs θ = α nd θ = β (Figure.3.7).

311 .3 Tngent Lines, Arc Length, nd Are for Polr Curves 73 u = u u = u n A u = u A n A Δu Δu u = b n u =..... Figure.3.8 u = u* k Δu k Figure.3.9 u = b Figure.3. Δu r = f (u) u* k u = In rectngulr coordintes we obtined res under curves b dividing the region into n incresing number of verticl strips, pproimting the strips b rectngles, nd tking limit. In polr coordintes rectngles re clums to work with, nd it is better to prtition the region into wedges b using rs such tht θ = θ,θ= θ,..., θ = θ n α<θ <θ < <θ n <β (Figure.3.8). As shown in tht figure, the rs divide the region R into n wedges with res A,A,...,A n nd centrl ngles θ, θ,..., θ n. The re of the entire region cn be written s n A = A + A + +A n = A k (4) If θ k is smll, then we cn pproimte the re A k of the kth wedge b the re of sector with centrl ngle θ k nd rdius f(θk ), where θ = θ k is n r tht lies in the kth wedge (Figure.3.9). Thus, from (4) nd Formul (5) of Appendi B for the re of sector, we obtin n n A = A k [f(θ k )] θ k (5) k= k= If we now increse n in such w tht m θ k, then the sectors will become better nd better pproimtions of the wedges nd it is resonble to epect tht (5) will pproch the ect vlue of the re A (Figure.3.); tht is, A = lim m θ k n k= [f(θ k )] θ k = k= β α [f(θ)] dθ Note tht the discussion bove cn esil be dpted to the cse where f(θ) is nonpositive for α θ β. We summrize this result below..3.4 re in polr coordintes If α nd β re ngles tht stisf the condition α<β α + π nd if f(θ)is continuous nd either nonnegtive or nonpositive for α θ β, then the re A of the region R enclosed b the polr curve r = f(θ)(α θ β) nd the lines θ = α nd θ = β is β β A = [f(θ)] dθ = r dθ (6) α α The hrdest prt of ppling (6) is determining the limits of integrtion. This cn be done s follows: Are in Polr Coordintes: Limits of Integrtion Step. Sketch the region R whose re is to be determined. Step. Drw n rbitrr rdil line from the pole to the boundr curve r = f(θ). Step 3. Ask, Over wht intervl of vlues must θ vr in order for the rdil line to sweep out the region R? Step 4. Your nswer in Step 3 will determine the lower nd upper limits of integrtion.

312 74 Chpter / Prmetric nd Polr Curves; Conic Sections Emple 6 r = cos θ. Find the re of the region in the first qudrnt tht is within the crdioid r = cos u c/ The shded region is swept out b the rdil line s u vries from to c/. Solution. The region nd tpicl rdil line re shown in Figure.3.. For the rdil line to sweep out the region, θ must vr from to π/. Thus, from (6) with α = nd β = π/, we obtin A = π/ r dθ = π/ ( cos θ) dθ = π/ ( cos θ + cos θ) dθ With the help of the identit cos θ = ( + cos θ), this cn be rewritten s A = π/ ( 3 cos θ + ) cos θ dθ = [ 3 θ sin θ + ] π/ sin θ = π Figure.3. Emple 7 Find the entire re within the crdioid of Emple 6. Solution. For the rdil line to sweep out the entire crdioid, θ must vr from to π. Thus, from (6) with α = nd β = π, π π A = r dθ = ( cos θ) dθ If we proceed s in Emple 6, this reduces to A = π ( 3 cos θ + ) cos θ dθ = 3π Alterntive Solution. Since the crdioid is smmetric bout the -is, we cn clculte the portion of the re bove the -is nd double the result. In the portion of the crdioid bove the -is, θ rnges from to π, so tht π π A = r dθ = ( cos θ) dθ = 3π USING SYMMETRY Although Formul (6) is pplicble if r = f(θ) is negtive, re computtions cn sometimes be simplified b using smmetr to restrict the limits of integrtion to intervls where r. This is illustrted in the net emple. Emple 8 Find the re of the region enclosed b the rose curve r = cos θ. Solution. Referring to Figure..3 nd using smmetr, the re in the first qudrnt tht is swept out for θ π/4 is one-eighth of the totl re inside the rose. Thus, from Formul (6) π/4 π/4 A = 8 r dθ = 4 cos θdθ = 4 π/4 ( + cos 4θ)dθ = = θ + ] π/4 sin 4θ = π π/4 ( + cos 4θ)dθ

313 .3 Tngent Lines, Arc Length, nd Are for Polr Curves 75 Sometimes the most nturl w to stisf the restriction α < β α + π required b Formul (6) is to use negtive vlue for α. For emple, suppose tht we re interested in finding the re of the shded region in Figure.3.. The first step would be to determine the intersections of the crdioid r = cos θ nd the circle r = 6, since this informtion is needed for the limits of integrtion. To find the points of intersection, we cn equte the two epressions for r. This ields which is stisfied b the positive ngles cos θ = 6 or cos θ = θ = π 3 nd θ = 5π 3 However, there is problem here becuse the rdil lines to the circle nd crdioid do not sweep through the shded region shown in Figure.3.b s θ vries over the intervl π/3 θ 5π/3. There re two ws to circumvent this problem one is to tke dvntge of the smmetr b integrting over the intervl θ π/3 nd doubling the result, nd the second is to use negtive lower limit of integrtion nd integrte over the intervl π/3 θ π/3 (Figure.3.c). The two methods re illustrted in the net emple. c/ c/ c/ c/ c/ r = 6 u = 4 u = 4 u = 4 u = 4 () r = cos u (b) u = k (c) u = $ (d) u = $ (e) u = $ Figure.3. Emple 9 Find the re of the region tht is inside of the crdioid r = cos θ nd outside of the circle r = 6. Solution Using Negtive Angle. The re of the region cn be obtined b subtrcting the res in Figures.3.d nd.3.e: A = = π/3 π/3 π/3 π/3 (4 + 4 cos θ) dθ π/3 π/3 [(4 + 4 cos θ) 36] dθ = (6) Are inside crdioid dθ minus re inside circle. π/3 π/3 = [ 6 sin θ + (4θ + sin θ) θ ] π/3 π/3 = 8 3 4π (6 cos θ + 8 cos θ )dθ Solution Using Smmetr. Using smmetr, we cn clculte the re bove the polr is nd double it. This ields (verif) A = π/3 which grees with the preceding result. [(4 + 4 cos θ) 36] dθ = (9 3 π) = 8 3 4π

314 76 Chpter / Prmetric nd Polr Curves; Conic Sections r = cos u Figure.3.3 c/ r = + cos u, 6, i The orbits intersect, but the stellites do not collide. Figure.3.4 INTERSECTIONS OF POLAR GRAPHS In the lst emple we found the intersections of the crdioid nd circle b equting their epressions for r nd solving for θ.however, becuse point cn be represented in different ws in polr coordintes, this procedure will not lws produce ll of the intersections. For emple, the crdioids r = cos θ nd r = + cos θ (7) intersect t three points: the pole, the point (,π/), nd the point (, 3π/) (Figure.3.3). Equting the right-hnd sides of the equtions in (7) ields cos θ = + cos θ or cos θ =, so θ = π + kπ, k =, ±, ±,... Substituting n of these vlues in (7) ields r =, so tht we hve found onl two distinct points of intersection, (,π/) nd (, 3π/); the pole hs been missed. This problem occurs becuse the two crdioids pss through the pole t different vlues of θ the crdioid r = cos θ psses through the pole t θ =, nd the crdioid r = + cos θ psses through the pole t θ = π. The sitution with the crdioids is nlogous to two stellites circling the Erth in intersecting orbits (Figure.3.4). The stellites will not collide unless the rech the sme point t the sme time. In generl, when looking for intersections of polr curves, it is good ide to grph the curves to determine how mn intersections there should be. QUICK CHECK EXERCISES.3 (See pge 79 for nswers.). () To obtin d/d directl from the polr eqution r = f(θ), we cn use the formul d d = d /dθ d/dθ = (b) Use the formul in prt () to find d/d directl from the polr eqution r = csc θ.. () Wht conditions on f(θ ) nd f (θ ) gurntee tht the line θ = θ is tngent to the polr curve r = f(θ)t the origin? (b) Wht re the vlues of θ in [, π] t which the lines θ = θ re tngent t the origin to the four-petl rose r = cos θ? 3. () To find the rc length L of the polr curve r = f(θ) (α θ β), we cn use the formul L =. (b) The polr curve r = sec θ( θ π/4) hs rc length L =. 4. The re of the region enclosed b nonnegtive polr curve r = f(θ)(α θ β) nd the lines θ = α nd θ = β is given b the definite integrl. 5. Find the re of the circle r = b integrtion. EXERCISE SET.3 Grphing Utilit C CAS 6 Find the slope of the tngent line to the polr curve for the given vlue of θ.. r = sin θ; θ = π/6. r = + cos θ; θ = π/ 3. r = /θ; θ = 4. r = sec θ; θ = π/6 5. r = sin 3θ; θ = π/4 6. r = 4 3 sin θ; θ = π c/ c/ 7 8 Clculte the slopes of the tngent lines indicted in the ccompning figures. 7. r = + sin θ 8. r = sin θ Figure E-7 Figure E-8 9 Find polr coordintes of ll points t which the polr curve hs horizontl or verticl tngent line. 9. r = ( + cos θ). r = sin θ

315 .3 Tngent Lines, Arc Length, nd Are for Polr Curves 77 Use grphing utilit to mke conjecture bout the number of points on the polr curve t which there is horizontl tngent line, nd confirm our conjecture b finding pproprite derivtives.. r = sin θ cos θ. r = sin θ 3 8 Sketch the polr curve nd find polr equtions of the tngent lines to the curve t the pole. 3. r = cos 3θ 4. r = 4 sin θ 5. r = 4 cos θ 6. r = sin θ 7. r = cos θ 8. r = θ 9 Use Formul (3) to clculte the rc length of the polr curve. 9. The entire circle r =. The entire circle r = cos θ. The entire crdioid r = ( cos θ). r = e 3θ from θ = toθ = 3. () Show tht the rc length of one petl of the rose r = cos nθ is given b π/(n) + (n ) sin nθ dθ (b) Use the numericl integrtion cpbilit of clculting utilit to pproimte the rc length of one petl of the four-petl rose r = cos θ. (c) Use the numericl integrtion cpbilit of clculting utilit to pproimte the rc length of one petl of the n-petl rose r = cos nθ for n =, 3, 4,...,; then mke conjecture bout the limit of these rc lengths s n () Sketch the spirl r = e θ/8 ( θ<+ ). (b) Find n improper integrl for the totl rc length of the spirl. (c) Show tht the integrl converges nd find the totl rc length of the spirl. 5. Write down, but do not evlute, n integrl for the re of ech shded region. () (b) (c) r = cos u r = cos u (d) (e) (f) r = u r = sin u r = sin u r = cos u 6. Find the re of the shded region in Eercise 5(d). 7. In ech prt, find the re of the circle b integrtion. () r = sin θ (b) r = cos θ 8. () Show tht r = sin θ + cos θ is circle. (b) Find the re of the circle using geometric formul nd then b integrtion Find the re of the region described. 9. The region tht is enclosed b the crdioid r = + sin θ. 3. The region in the first qudrnt within the crdioid r = + cos θ. 3. The region enclosed b the rose r = 4 cos 3θ. 3. The region enclosed b the rose r = sin θ. 33. The region enclosed b the inner loop of the limçon r = + cos θ. [Hint: r over the intervl of integrtion.] 34. The region swept out b rdil line from the pole to the curve r = /θ s θ vries over the intervl θ Find the re of the shded region r = cos u r = cos u r = 4 cos u r = 4 3 sin u 38. r = + cos u r = cos u r = + cos u r = 3 cos u Find the re of the region described. 39. The region inside the circle r = 3 sin θ nd outside the crdioid r = + sin θ. 4. The region outside the crdioid r = cos θ nd inside the circle r = The region inside the crdioid r = + cos θ nd outside the circle r = The region tht is common to the circles r = cos θ nd r = sin θ. 43. The region between the loops of the limçon r = + cos θ. 44. The region inside the crdioid r = + cos θ nd to the right of the line r cos θ = The region inside the circle r = nd to the right of the line r = sec θ. 46. The region inside the rose r = cos θ nd outside the circle r =.

316 78 Chpter / Prmetric nd Polr Curves; Conic Sections 47 5 True Flse Determine whether the sttement is true or flse. Eplin our nswer. 47. The -is is tngent to the polr curve r = cos(θ/) t θ = 3π. 48. The rc length of the polr curve r = θ for θ π/ is given b π/ L = + 4θ dθ Bug t = s 49. The re of sector with centrl ngle θ tken from circle of rdius r is θr. 5. The epression π/4 π/4 ( cos θ) dθ computes the re enclosed b the inner loop of the limçon r = cos θ. FOCUS ON CONCEPTS 5. () Find the error: The re tht is inside the lemniscte r = cos θ is A = π = 4 sin θ π r dθ = cos θ dθ ] π = (b) Find the correct re. (c) Find the re inside the lemniscte r = 4 cos θ nd outside the circle r =. 5. Find the re inside the curve r = sin θ. 53. A rdil line is drwn from the origin to the spirl r = θ ( > nd θ ). Find the re swept out during the second revolution of the rdil line tht ws not swept out during the first revolution. 54. As illustrted in the ccompning figure, suppose tht rod with one end fied t the pole of polr coordinte sstem rottes counterclockwise t the constnt rte of rd/s. At time t = bug on the rod is mm from the pole nd is moving outwrd long the rod t the constnt speed of mm/s. () Find n eqution of the form r = f(θ) for the pth of motion of the bug, ssuming tht θ = when t =. (b) Find the distnce the bug trvels long the pth in prt () during the first 5 s. Round our nswer to the nerest tenth of millimeter. t = 5 s Figure E-54 C 55. () Show tht the Folium of Descrtes = cn be epressed in polr coordintes s 3 sin θ cos θ r = cos 3 θ + sin 3 θ (b) Use CAS to show tht the re inside of the loop is 3 (Figure 3..3). C 56. () Wht is the re tht is enclosed b one petl of the rose r = cos nθ if n is n even integer? (b) Wht is the re tht is enclosed b one petl of the rose r = cos nθ if n is n odd integer? (c) Use CAS to show tht the totl re enclosed b the rose r = cos nθ is π / if the number of petls is even. [Hint: See Eercise 78 of Section..] (d) Use CAS to show tht the totl re enclosed b the rose r = cos nθ is π /4 if the number of petls is odd. 57. One of the most fmous problems in Greek ntiquit ws squring the circle, tht is, using strightedge nd compss to construct squre whose re is equl to tht of given circle. It ws proved in the nineteenth centur tht no such construction is possible. However, show tht the shded res in the ccompning figure re equl, thereb squring the crescent. c/ Figure E Use grphing utilit to generte the polr grph of the eqution r = cos 3θ +, nd find the re tht it encloses. 59. Use grphing utilit to generte the grph of the bifolium r = cos θ sin θ, nd find the re of the upper loop. 6. Use Formul (9) of Section. to derive the rc length formul for polr curves, Formul (3). 6. As illustrted in the ccompning figure, let P(r,θ) be point on the polr curve r = f(θ), let ψ be the smllest counterclockwise ngle from the etended rdius OP to the

317 .3 Tngent Lines, Arc Length, nd Are for Polr Curves 79 tngent line t P, nd let φ be the ngle of inclintion of the tngent line. Derive the formul tn ψ = r dr/dθ b substituting tn φ for d/d in Formul () nd ppling the trigonometric identit tn φ tn θ tn(φ θ) = + tn φ tn θ O r = f (u) u c P(r, u) Tngent line f Figure E Use the formul for ψ obtined in Eercise () Use the trigonometric identit tn θ = cos θ sin θ to show tht if (r, θ) is point on the crdioid r = cos θ ( θ<π) then ψ = θ/. (b) Sketch the crdioid nd show the ngle ψ t the points where the crdioid crosses the -is. (c) Find the ngle ψ t the points where the crdioid crosses the -is. 63. Show tht for logrithmic spirl r = e bθ, the ngle from the rdil line to the tngent line is constnt long the spirl (see the ccompning figure). [Note: For this reson, logrithmic spirls re sometimes clled equingulr spirls.] Figure E () In the discussion ssocited with Eercises 75 8 of Section., formuls were given for the re of the surfce of revolution tht is generted b revolving prmetric curve bout the -is or -is. Use those formuls to derive the following formuls for the res of the surfces of revolution tht re generted b revolving the portion of the polr curve r = f(θ) from θ = α to θ = β bout the polr is nd bout the line θ = π/: β ( ) dr S = πr sin θ r + dθ About θ = α dθ β ( ) dr S = πr cos θ r + dθ About θ = π/ dθ α (b) Stte conditions under which these formuls hold Sketch the surfce, nd use the formuls in Eercise 64 to find the surfce re. 65. The surfce generted b revolving the circle r = cos θ bout the line θ = π/. 66. The surfce generted b revolving the spirl r = e θ ( θ π/) bout the line θ = π/. 67. The pple generted b revolving the upper hlf of the crdioid r = cos θ( θ π) bout the polr is. 68. The sphere of rdius generted b revolving the semicircle r = in the upper hlf-plne bout the polr is. 69. Writing () Show tht if θ <θ π nd if r nd r re positive, then the re A of tringle with vertices (, ), (r,θ ), nd (r,θ ) is A = r r sin(θ θ ) (b) Use the formul obtined in prt () to describe n pproch to nswer Are Problem.3.3 tht uses n pproimtion of the region R b tringles insted of circulr wedges. Reconcile our pproch with Formul (6). 7. Writing In order to find the re of region bounded b two polr curves it is often necessr to determine their points of intersection. Give n emple to illustrte tht the points of intersection of curves r = f(θ) nd r = g(θ) m not coincide with solutions to f(θ)= g(θ). Discuss some strtegies for determining intersection points of polr curves nd provide emples to illustrte our strtegies. QUICK CHECK ANSWERS.3. () 3. () r cos θ + sin θ dr dθ r sin θ + cos θ dr dθ β α r + (b) d d =. () f(θ ) =, f (θ ) = (b) θ = π 4, 3π 4, 5π 4, 7π 4 ( ) dr dθ (b) 4. dθ β α β [f(θ)] dθ = r dθ 5. α π dθ = π

318 73 Chpter / Prmetric nd Polr Curves; Conic Sections.4 CONIC SECTIONS In this section we will discuss some of the bsic geometric properties of prbols, ellipses, nd hperbols. These curves pl n importnt role in clculus nd lso rise nturll in brod rnge of pplictions in such fields s plnetr motion, design of telescopes nd ntenns, geodetic positioning, nd medicine, to nme few. CONIC SECTIONS Circles, ellipses, prbols, nd hperbols re clled conic sections or conics becuse the cn be obtined s intersections of plne with double-npped circulr cone (Figure.4.). If the plne psses through the verte of the double-npped cone, then the intersection is point, pir of intersecting lines, or single line. These re clled degenerte conic sections. Circle Ellipse Prbol Hperbol Figure.4. A point A pir of intersecting lines A single line Some students m lred be fmilir with the mteril in this section, in which cse it cn be treted s review. Instructors who wnt to spend some dditionl time on preclculus review m wnt to llocte more thn one lecture on this mteril.

319 .4 Conic Sections 73 DEFINITIONS OF THE CONIC SECTIONS Although we could derive properties of prbols, ellipses, nd hperbols b defining them s intersections with double-npped cone, it will be better suited to clculus if we begin with equivlent definitions tht re bsed on their geometric properties..4. definition A prbol is the set of ll points in the plne tht re equidistnt from fied line nd fied point not on the line. All points on the prbol re equidistnt from the focus nd directri. The line is clled the directri of the prbol, nd the point is clled the focus (Figure.4.). A prbol is smmetric bout the line tht psses through the focus t right ngles to the directri. This line, clled the is or the is of smmetr of the prbol, intersects the prbol t point clled the verte. Focus Verte Directri Figure.4. Ais.4. definition An ellipse is the set of ll points in the plne, the sum of whose distnces from two fied points is given positive constnt tht is greter thn the distnce between the fied points. The two fied points re clled the foci (plurl of focus ) of the ellipse, nd the midpoint of the line segment joining the foci is clled the center (Figure.4.3). To help visulize Definition.4., imgine tht two ends of string re tcked to the foci nd pencil trces curve s it is held tight ginst the string (Figure.4.3b). The resulting curve will be n ellipse since the sum of the distnces to the foci is constnt, nmel, the totl length of the string. Note tht if the foci coincide, the ellipse reduces to circle. For ellipses other thn circles, the line segment through the foci nd cross the ellipse is clled the mjor is (Figure.4.3c), nd the line segment cross the ellipse, through the center, nd perpendiculr to the mjor is is clled the minor is. The endpoints of the mjor is re clled vertices. The sum of the distnces to the foci is constnt. Mjor is Minor is Verte Verte Focus Center Focus Figure.4.3 () (b) (c).4.3 definition A hperbol is the set of ll points in the plne, the difference of whose distnces from two fied distinct points is given positive constnt tht is less thn the distnce between the fied points. The two fied points re clled the foci of the hperbol, nd the term difference tht is used in the definition is understood to men the distnce to the frther focus minus the distnce to the closer focus. As result, the points on the hperbol form two brnches, ech

320 73 Chpter / Prmetric nd Polr Curves; Conic Sections wrpping round the closer focus (Figure.4.4). The midpoint of the line segment joining the foci is clled the center of the hperbol, the line through the foci is clled the focl is, nd the line through the center tht is perpendiculr to the focl is is clled the conjugte is. The hperbol intersects the focl is t two points clled the vertices. Associted with ever hperbol is pir of lines, clled the smptotes of the hperbol. These lines intersect t the center of the hperbol nd hve the propert tht s point P moves long the hperbol w from the center, the verticl distnce between P nd one of the smptotes pproches zero (Figure.4.4b). Conjugte is The distnce from the frther focus minus the distnce to the closer focus is constnt. Focus Center Verte Verte Focl is Focus These distnces pproch zero s the point moves w from the center. These distnces pproch zero s the point moves w from the center. Figure.4.4 () (b) p p Directri Figure.4.5 p p Ais EQUATIONS OF PARABOLAS IN STANDARD POSITION It is trditionl in the stud of prbols to denote the distnce between the focus nd the verte b p. The verte is equidistnt from the focus nd the directri, so the distnce between the verte nd the directri is lso p; consequentl, the distnce between the focus nd the directri is p (Figure.4.5). As illustrted in tht figure, the prbol psses through two of the corners of bo tht etends from the verte to the focus long the is of smmetr nd etends p units bove nd p units below the is of smmetr. The eqution of prbol is simplest if the verte is the origin nd the is of smmetr is long the -is or -is. The four possible such orienttions re shown in Figure.4.6. These re clled the stndrd positions of prbol, nd the resulting equtions re clled the stndrd equtions of prbol. prbols in stndrd position = p ( p, ) ( p, ) (, p) (, p) = p = p = p = 4p = 4p = 4p = 4p Figure.4.6

321 D( p, ) P(, ) F(p, ) = p Figure Conic Sections 733 To illustrte how the equtions in Figure.4.6 re obtined, we will derive the eqution for the prbol with focus (p, ) nd directri = p. Let P(,) be n point on the prbol. Since P is equidistnt from the focus nd directri, the distnces PF nd PD in Figure.4.7 re equl; tht is, PF = PD () where D( p, ) is the foot of the perpendiculr from P to the directri. From the distnce formul, the distnces PF nd PD re Substituting in () nd squring ields nd fter simplifing PF = ( p) + nd PD = ( + p) () ( p) + = ( + p) (3) = 4p (4) The derivtions of the other equtions in Figure.4.6 re similr. A TECHNIQUE FOR SKETCHING PARABOLAS Prbols cn be sketched from their stndrd equtions using four bsic steps: Figure.4.8 Rough sketch Sketching Prbol from Its Stndrd Eqution Step. Determine whether the is of smmetr is long the -is or the -is. Referring to Figure.4.6, the is of smmetr is long the -is if the eqution hs -term, nd it is long the -is if it hs n -term. Step. Determine which w the prbol opens. If the is of smmetr is long the -is, then the prbol opens to the right if the coefficient of is positive, nd it opens to the left if the coefficient is negtive. If the is of smmetr is long the -is, then the prbol opens up if the coefficient of is positive, nd it opens down if the coefficient is negtive. Step 3. Determine the vlue of p nd drw bo etending p units from the origin long the is of smmetr in the direction in which the prbol opens nd etending p units on ech side of the is of smmetr. Step 4. Using the bo s guide, sketch the prbol so tht its verte is t the origin nd it psses through the corners of the bo (Figure.4.8). (, 3) 6 = Figure = 3 Emple Sketch the grphs of the prbols nd show the focus nd directri of ech. () = (b) + 8 = Solution (). This eqution involves, so the is of smmetr is long the -is, nd the coefficient of is positive, so the prbol opens upwrd. From the coefficient of, we obtin 4p = or p = 3. Drwing bo etending p = 3 units up from the origin nd p = 6 units to the left nd p = 6 units to the right of the -is, then using corners of the bo s guide, ields the grph in Figure.4.9. The focus is p = 3 units from the verte long the is of smmetr in the direction in which the prbol opens, so its coordintes re (, 3). The directri is perpendiculr to the is of smmetr t distnce of p = 3 units from the verte on the opposite side from the focus, so its eqution is = 3.

322 734 Chpter / Prmetric nd Polr Curves; Conic Sections Solution (b). We first rewrite the eqution in the stndrd form = 8 (, ) 4 This eqution involves, so the is of smmetr is long the -is, nd the coefficient of is negtive, so the prbol opens to the left. From the coefficient of we obtin 4p = 8, so p =. Drwing bo etending p = units left from the origin nd p = 4 units bove nd p = 4 units below the -is, then using corners of the bo s guide, ields the grph in Figure.4.. Figure.4. 4 = 8 = Emple Find n eqution of the prbol tht is smmetric bout the -is, hs its verte t the origin, nd psses through the point (5, ). Solution. Since the prbol is smmetric bout the -is nd hs its verte t the origin, the eqution is of the form = 4p or = 4p where the sign depends on whether the prbol opens up or down. But the prbol must open up since it psses through the point (5, ), which lies in the first qudrnt. Thus, the eqution is of the form = 4p (5) Since the prbol psses through (5, ), we must hve 5 = 4p or4p = 5. Therefore, (5) becomes = 5 c Figure.4. c b b EQUATIONS OF ELLIPSES IN STANDARD POSITION It is trditionl in the stud of ellipses to denote the length of the mjor is b, the length of the minor is b b, nd the distnce between the foci b c (Figure.4.). The number is clled the semimjor is nd the number b the semiminor is (stndrd but odd terminolog, since nd b re numbers, not geometric es). There is bsic reltionship between the numbers, b, nd c tht cn be obtined b emining the sum of the distnces to the foci from point P t the end of the mjor is nd from point Q t the end of the minor is (Figure.4.). From Definition.4., these sums must be equl, so we obtin b + c Q b c Figure.4. b Figure.4.3 b + c c P c c from which it follows tht or, equivlentl, b + c = ( c) + ( + c) = b + c (6) c = b (7) From (6), the distnce from focus to n end of the minor is is (Figure.4.3), which implies tht for ll points on the ellipse the sum of the distnces to the foci is. It lso follows from (6) tht b with the equlit holding onl when c =. Geometricll, this mens tht the mjor is of n ellipse is t lest s lrge s the minor is nd tht the two es hve equl length onl when the foci coincide, in which cse the ellipse is circle. The eqution of n ellipse is simplest if the center of the ellipse is t the origin nd the foci re on the -is or -is. The two possible such orienttions re shown in Figure.4.4.

323 .4 Conic Sections 735 These re clled the stndrd positions of n ellipse, nd the resulting equtions re clled the stndrd equtions of n ellipse. ellipses in stndrd position b (, c) ( c, ) (c, ) b b b (, c) + = b + = b Figure.4.4 To illustrte how the equtions in Figure.4.4 re obtined, we will derive the eqution for the ellipse with foci on the -is. Let P(,) be n point on tht ellipse. Since the sum of the distnces from P to the foci is, it follows (Figure.4.5) tht P(, ) PF + PF = F ( c, ) F(c, ) so ( + c) + + ( c) + = Figure.4.5 Trnsposing the second rdicl to the right side of the eqution nd squring ields ( + c) + = 4 4 ( c) + + ( c) + nd, on simplifing, ( c) + = c (8) Squring gin nd simplifing ields which, b virtue of (6), cn be written s + c = + b = (9) Conversel, it cn be shown tht n point whose coordintes stisf (9) hs s the sum of its distnces from the foci, so tht such point is on the ellipse.

324 736 Chpter / Prmetric nd Polr Curves; Conic Sections A TECHNIQUE FOR SKETCHING ELLIPSES Ellipses cn be sketched from their stndrd equtions using three bsic steps: Figure.4.6 Rough sketch Sketching n Ellipse from Its Stndrd Eqution Step. Determine whether the mjor is is on the -is or the -is. This cn be scertined from the sizes of the denomintors in the eqution. Referring to Figure.4.4, nd keeping in mind tht >b (since >b), the mjor is is long the -is if hs the lrger denomintor, nd it is long the -is if hs the lrger denomintor. If the denomintors re equl, the ellipse is circle. Step. Determine the vlues of nd b nd drw bo etending units on ech side of the center long the mjor is nd b units on ech side of the center long the minor is. Step 3. Using the bo s guide, sketch the ellipse so tht its center is t the origin nd it touches the sides of the bo where the sides intersect the coordinte es (Figure.4.6). 4 (, 7) Emple 3 Sketch the grphs of the ellipses () = (b) + = 4 showing the foci of ech Figure (, 7) + = 6 Solution (). Since hs the lrger denomintor, the mjor is is long the -is. Moreover, since >b, we must hve = 6 nd b = 9, so = 4 nd b = 3 Drwing bo etending 4 units on ech side of the origin long the -is nd 3 units on ech side of the origin long the -is s guide ields the grph in Figure.4.7. The foci lie c units on ech side of the center long the mjor is, where c is given b (7). From the vlues of nd b bove, we obtin c = b = 6 9 = 7.6 Thus, the coordintes of the foci re (, 7 ) nd (, 7 ), since the lie on the -is. (, ) 4 Figure = (, ) Solution (b). We first rewrite the eqution in the stndrd form 4 + = Since hs the lrger denomintor, the mjor is lies long the -is, nd we hve = 4 nd b =. Drwing bo etending = units on ech side of the origin long the -is nd etending b =.4 units on ech side of the origin long the -is s guide ields the grph in Figure.4.8. From (7), we obtin c = b =.4 Thus, the coordintes of the foci re (, ) nd (, ), since the lie on the -is.

325 .4 Conic Sections 737 Find n eqution for the ellipse with foci (, ±) nd mjor is with end- Emple 4 points (, ±4). Solution. From Figure.4.4, the eqution hs the form b + = nd from the given informtion, = 4 nd c =. It follows from (6) tht so the eqution of the ellipse is b = c = 6 4 = + 6 = c c EQUATIONS OF HYPERBOLAS IN STANDARD POSITION It is trditionl in the stud of hperbols to denote the distnce between the vertices b, the distnce between the foci b c (Figure.4.9), nd to define the quntit b s b = c () Figure.4.9 b Figure.4. c Figure.4. c V c This reltionship, which cn lso be epressed s c = + b () is pictured geometricll in Figure.4.. As illustrted in tht figure, nd s we will show lter in this section, the smptotes pss through the corners of bo etending b units on ech side of the center long the conjugte is nd units on ech side of the center long the focl is. The number is clled the semifocl is of the hperbol nd the number b the semiconjugte is. (As with the semimjor nd semiminor es of n ellipse, these re numbers, not geometric es.) If V is one verte of hperbol, then, s illustrted in Figure.4., the distnce from V to the frther focus minus the distnce from V to the closer focus is [(c ) + ] (c ) = Thus, for ll points on hperbol, the distnce to the frther focus minus the distnce to the closer focus is. The eqution of hperbol hs n especill convenient form if the center of the hperbol is t the origin nd the foci re on the -is or -is. The two possible such orienttions re shown in Figure.4.. These re clled the stndrd positions of hperbol, nd the resulting equtions re clled the stndrd equtions of hperbol. The derivtions of these equtions re similr to those lred given for prbols nd ellipses, so we will leve them s eercises. However, to illustrte how the equtions of the smptotes re derived, we will derive those equtions for the hperbol We cn rewrite this eqution s b = = b ( ) which is equivlent to the pir of equtions = b nd = b

326 738 Chpter / Prmetric nd Polr Curves; Conic Sections hperbols in stndrd position = b (, c) = b b ( c, ) (c, ) b b b = b (, c) = b Figure.4. b = b = Figure.4.3 = b = b Thus, in the first qudrnt, the verticl distnce between the line = (b/) nd the hperbol cn be written s b b (Figure.4.3). But this distnce tends to zero s + since ( b lim + b ) b = lim ( ) + ( )( + ) b = lim + + b = lim + + = The nlsis in the remining qudrnts is similr. A QUICK WAY TO FIND ASYMPTOTES There is trick tht cn be used to void memorizing the equtions of the smptotes of hperbol. The cn be obtined, when needed, b replcing b on the right side of the hperbol eqution, nd then solving for in terms of. For emple, for the hperbol we would write b = b = or = b or =± b which re the equtions for the smptotes.

327 .4 Conic Sections 739 A TECHNIQUE FOR SKETCHING HYPERBOLAS Hperbols cn be sketched from their stndrd equtions using four bsic steps: Figure.4.4 Rough sketch Sketching Hperbol from Its Stndrd Eqution Step. Determine whether the focl is is on the -is or the -is. This cn be scertined from the loction of the minus sign in the eqution. Referring to Figure.4., the focl is is long the -is when the minus sign precedes the -term, nd it is long the -is when the minus sign precedes the -term. Step. Determine the vlues of nd b nd drw bo etending units on either side of the center long the focl is nd b units on either side of the center long the conjugte is. (The squres of nd b cn be red directl from the eqution.) Step 3. Drw the smptotes long the digonls of the bo. Step 4. Using the bo nd the smptotes s guide, sketch the grph of the hperbol (Figure.4.4). Emple 5 Sketch the grphs of the hperbols = 3 5 = 3 () 4 9 = (b) = showing their vertices, foci, nd smptotes. 3 Figure.4.5 = = = Solution (). The minus sign precedes the -term, so the focl is is long the -is. From the denomintors in the eqution we obtin = 4 nd b = 9 Since nd b re positive, we must hve = nd b = 3. Reclling tht the vertices lie units on ech side of the center on the focl is, it follows tht their coordintes in this cse re (, ) nd (, ). Drwing bo etending = units long the -is on ech side of the origin nd b = 3 units on ech side of the origin long the -is, then drwing the smptotes long the digonls of the bo s guide, ields the grph in Figure.4.5. To obtin equtions for the smptotes, we replce b in the given eqution; this ields 4 9 = or =±3 The foci lie c units on ech side of the center long the focl is, where c is given b (). From the vlues of nd b bove we obtin c = + b = = Since the foci lie on the -is in this cse, their coordintes re ( 3, ) nd ( 3, ) Figure.4.6 Solution (b). The minus sign precedes the -term, so the focl is is long the -is. From the denomintors in the eqution we obtin = nd b =, from which it follows tht = nd b = Thus, the vertices re t (, ) nd (, ). Drwing bo etending = unit on either side of the origin long the -is nd b = unit on either side of the origin long the -is, then drwing the smptotes, ields the grph in Figure.4.6. Since the bo is ctull

328 74 Chpter / Prmetric nd Polr Curves; Conic Sections A hperbol in which = b, s in prt (b) of Emple 5, is clled n equilterl hperbol. Such hperbols lws hve perpendiculr smptotes. squre, the smptotes re perpendiculr nd hve equtions =±. This cn lso be seen b replcing b in the given eqution, which ields = or =±.Also, c = + b = + = so the foci, which lie on the -is, re (, ) nd (, ). Emple 6 =± 4 3. Find the eqution of the hperbol with vertices (, ±8) nd smptotes Solution. Since the vertices re on the -is, the eqution of the hperbol hs the form ( / ) ( /b ) = nd the smptotes re =± b From the loctions of the vertices we hve = 8, so the given equtions of the smptotes ield =± b =±8 b =±4 3 from which it follows tht b = 6. Thus, the hperbol hs the eqution = TRANSLATED CONICS Equtions of conics tht re trnslted from their stndrd positions cn be obtined b replcing b h nd b k in their stndrd equtions. For prbol, this trnsltes the verte from the origin to the point (h, k); nd for ellipses nd hperbols, this trnsltes the center from the origin to the point (h, k). Prbols with verte (h, k) nd is prllel to -is ( k) = 4p( h) [Opens right] () ( k) = 4p( h) [Opens left] (3) Prbols with verte (h, k) nd is prllel to -is ( h) = 4p( k) [Opens up] (4) ( h) = 4p( k) [Opens down] (5) Ellipse with center (h, k) nd mjor is prllel to -is ( h) ( k) + = [b <] (6) b Ellipse with center (h, k) nd mjor is prllel to -is ( h) ( k) + = [b <] (7) b Hperbol with center (h, k) nd focl is prllel to -is ( h) ( k) = (8) b Hperbol with center (h, k) nd focl is prllel to -is ( k) ( h) b = (9)

329 .4 Conic Sections 74 Emple 7 t (4, ). Find n eqution for the prbol tht hs its verte t (, ) nd its focus Solution. Since the focus nd verte re on horizontl line, nd since the focus is to the right of the verte, the prbol opens to the right nd its eqution hs the form ( k) = 4p( h) Since the verte nd focus re 3 units prt, we hve p = 3, nd since the verte is t (h, k) = (, ), we obtin ( ) = ( ) Sometimes the equtions of trnslted conics occur in epnded form, in which cse we re fced with the problem of identifing the grph of qudrtic eqution in nd : A + C + D + E + F = () The bsic procedure for determining the nture of such grph is to complete the squres of the qudrtic terms nd then tr to mtch up the resulting eqution with one of the forms of trnslted conic. Emple 8 Describe the grph of the eqution = ( 4, 3) Directri = 6 Figure.4.7 (, 3) = Solution. The eqution involves qudrtic terms in but none in, so we first tke ll of the -terms to one side: 6 = Net, we complete the squre on the -terms b dding 9 to both sides: ( 3) = Finll, we fctor out the coefficient of the -term to obtin ( 3) = 8( + 4) This eqution is of form () with h = 4,k = 3, nd p =, so the grph is prbol with verte ( 4, 3) opening to the right. Since p =, the focus is units to the right of the verte, which plces it t the point (, 3); nd the directri is units to the left of the verte, which mens tht its eqution is = 6. The prbol is shown in Figure.4.7. Emple 9 Describe the grph of the eqution = Solution. This eqution involves qudrtic terms in both nd, so we will group the -terms nd the -terms on one side nd put the constnt on the other: (6 64) + (9 54) = Net, fctor out the coefficients of nd nd complete the squres: 6( 4 + 4) + 9( 6 + 9) = or 6( ) + 9( 3) = 44

330 74 Chpter / Prmetric nd Polr Curves; Conic Sections (, 7) (, 3 + 7) (, 3) (, 3) (5, 3) (, 3 7) (, ) = Figure.4.8 Finll, divide through b 44 to introduce ontheright side: ( ) + 9 ( 3) 6 This is n eqution of form (7), with h =, k = 3, = 6, nd b = 9. Thus, the grph of the eqution is n ellipse with center (, 3) nd mjor is prllel to the -is. Since = 4, the mjor is etends 4 units bove nd 4 units below the center, so its endpoints re (, 7) nd (, ) (Figure.4.8). Since b = 3, the minor is etends 3 units to the left nd 3 units to the right of the center, so its endpoints re (, 3) nd (5, 3). Since = c = b = 6 9 = 7 the foci lie 7 units bove nd below the center, plcing them t the points (, ) nd (, 3 7 ). Emple Describe the grph of the eqution = = + 6 = + ( 3, 4) ( + 3, 4) = Figure.4.9 Solution. This eqution involves qudrtic terms in both nd, so we will group the -terms nd the -terms on one side nd put the constnt on the other: ( 4) ( 8) = We leve it for ou to verif b completing the squres tht this eqution cn be written s ( ) 9 ( 4) 9 = () This is n eqution of form (8) with h =,k= 4, = 9, nd b = 9. Thus, the eqution represents hperbol with center (, 4) nd focl is prllel to the -is. Since = 3, the vertices re locted 3 units to the left nd 3 units to the right of the center, or t the points (, 4) nd (5, 4). From (), c = + b = = 3, so the foci re locted 3 units to the left nd right of the center, or t the points ( 3, 4) nd ( + 3, 4). The equtions of the smptotes m be found using the trick of replcing b in () to obtin ( ) ( 4) = 9 9 This cn be written s 4 =±( ), which ields the smptotes = + nd = + 6 With the id of bo etending = 3 units left nd right of the center nd b = 3 units bove nd below the center, we obtin the sketch in Figure.4.9. REFLECTION PROPERTIES OF THE CONIC SECTIONS Prbols, ellipses, nd hperbols hve certin reflection properties tht mke them etremel vluble in vrious pplictions. In the eercises we will sk ou to prove the following results..4.4 theorem (Reflection Propert of Prbols) The tngent line t point P on prbol mkes equl ngles with the line through P prllel to the is of smmetr nd the line through P nd the focus (Figure.4.3).

331 .4 Conic Sections theorem (Reflection Propert of Ellipses) A line tngent to n ellipse t point P mkes equl ngles with the lines joining P to the foci (Figure.4.3b)..4.6 theorem (Reflection Propert of Hperbols) A line tngent to hperbol t point P mkes equl ngles with the lines joining P to the foci (Figure.4.3c). Figure.4.3 P Ais of smmetr Tngent line t P P Focus P Tngent line t P Tngent line t P () (b) (c) John Med/Science Photo Librr/Photo Reserchers Incoming signls re reflected b the prbolic ntenn to the receiver t the focus. APPLICATIONS OF THE CONIC SECTIONS Fermt s principle in optics implies tht light reflects off of surfce t n ngle equl to its ngle of incidence. (See Eercise 64 in Section 4.5.) In prticulr, if reflecting surfce is generted b revolving prbol bout its is of smmetr, it follows from Theorem.4.4 tht ll light rs entering prllel to the is will be reflected to the focus (Figure.4.3); conversel, if light source is locted t the focus, then the reflected rs will ll be prllel to the is (Figure.4.3b). This principle is used in certin telescopes to reflect the pproimtel prllel rs of light from the strs nd plnets off of prbolic mirror to n eepiece t the focus; nd the prbolic reflectors in flshlights nd utomobile hedlights utilize this principle to form prllel bem of light rs from bulb plced t the focus. The sme opticl principles ppl to rdr signls nd sound wves, which eplins the prbolic shpe of mn ntenns. Figure.4.3 () (b) Visitors to vrious rooms in the United Sttes Cpitol Building nd in St. Pul s Cthedrl in London re often stonished b the whispering gller effect in which two people t opposite ends of the room cn her one nother s whispers ver clerl. Such rooms hve ceilings with ellipticl cross sections nd common foci. Thus, when the two people stnd t the foci, their whispers re reflected directl to one nother off of the ellipticl ceiling. Hperbolic nvigtion sstems, which were developed in World Wr II s nvigtionl ids to ships, re bsed on the definition of hperbol. With these sstems the ship receives

332 744 Chpter / Prmetric nd Polr Curves; Conic Sections Ship snchronized rdio signls from two widel spced trnsmitters with known positions. The ship s electronic receiver mesures the difference in reception times between the signls nd then uses tht difference to compute the difference between its distnces from the two trnsmitters. This informtion plces the ship somewhere on the hperbol whose foci re t the trnsmitters nd whose points hve s the difference in their distnces from the foci. B repeting the process with second set of trnsmitters, the position of the ship cn be pproimted s the intersection of two hperbols (Figure.4.3). [The modern globl positioning sstem (GPS) is bsed on the sme principle.] Atlntic Ocen Figure.4.3 QUICK CHECK EXERCISES.4 (See pge 748 for nswers.). Identif the conic. () The set of points in the plne, the sum of whose distnces to two fied points is positive constnt greter thn the distnce between the fied points is. (b) The set of points in the plne, the difference of whose distnces to two fied points is positive constnt less thn the distnce between the fied points is. (c) The set of points in the plne tht re equidistnt from fied line nd fied point not on the line is.. () The eqution of the prbol with focus (p, ) nd directri = p is. (b) The eqution of the prbol with focus (,p)nd directri = p is. 3. () Suppose tht n ellipse hs semimjor is nd semiminor is b. Then for ll points on the ellipse, the sum of the distnces to the foci is equl to. (b) The two stndrd equtions of n ellipse with semimjor is nd semiminor is b re nd. (c) Suppose tht n ellipse hs semimjor is, semiminor is b, nd foci (±c, ). Then c m be obtined from nd b b the eqution c =. 4. () Suppose tht hperbol hs semifocl is nd semiconjugte is b. Then for ll points on the hperbol, the difference of the distnce to the frther focus minus the distnce to the closer focus is equl to. (b) The two stndrd equtions of hperbol with semifocl is nd semiconjugte is b re nd. (c) Suppose tht hperbol in stndrd position hs semifocl is, semiconjugte is b, nd foci (±c, ). Then c m be obtined from nd b b the eqution c =. The equtions of the smptotes of this hperbol re =±. EXERCISE SET.4 Grphing Utilit C CAS FOCUS ON CONCEPTS. In prts () (f), find the eqution of the conic. () (b) (c) 3 (d)

333 .4 Conic Sections 745 (e) (f) () Find the focus nd directri for ech prbol in Eercise. (b) Find the foci of the ellipses in Eercise. (c) Find the foci nd the equtions of the smptotes of the hperbols in Eercise. 3 6 Sketch the prbol, nd lbel the focus, verte, nd directri. 3. () = 4 (b) = 8 4. () = (b) = 4 5. () ( ) = ( + 4) (b) ( ) = ( 6. () 6 + = (b) = Sketch the ellipse, nd lbel the foci, vertices, nd ends of the minor is. 7. () = (b) 9 + = 9 8. () = (b) 4 + = () ( + 3) + 4( 5) = 6 (b) ( + ) =. () = (b) = 56 4 Sketch the hperbol, nd lbel the vertices, foci, nd smptotes.. () 6 9 = (b) 9 = 36. () 9 5 = (b) 6 5 = 4 ( + 4) ( ) 3. () = 3 5 (b) 6( + ) 8( 3) = 6 4. () = (b) = Find n eqution for the prbol tht stisfies the given conditions. 5. () Verte (, ); focus (3, ). (b) Verte (, ); directri = () Focus (6, ); directri = 6. (b) Focus (, ); directri =. 7. Ais = ; psses through (3, ) nd (, ). ) 8. Verte (5, 3); is prllel to the -is; psses through (9, 5). 9 Find n eqution for the ellipse tht stisfies the given conditions. 9. () Ends of mjor is (±3, ); ends of minor is (, ±). (b) Length of minor is 8; foci (, ±3).. () Foci (±, ); b =. (b) c = 3; = 4; center t the origin; foci on coordinte is (two nswers).. () Ends of mjor is (, ±6); psses through ( 3, ). (b) Foci (, ) nd (, 3); minor is of length 4.. () Center t (, ); mjor nd minor es long the coordinte es; psses through (3, ) nd (, 6). (b) Foci (, ) nd (, 3); mjor is of length Find n eqution for hperbol tht stisfies the given conditions. [Note: In some cses there m be more thn one hperbol.] 3. () Vertices (±, ); foci (±3, ). (b) Vertices (, ±); smptotes =± () Asmptotes =± 3 ; b = 4. (b) Foci (, ±5); smptotes =±. 5. () Asmptotes =± 3 ; c = 5. 4 (b) Foci (±3, ); smptotes =±. 6. () Vertices (, 6) nd (6, 6); foci units prt. (b) Asmptotes = nd = + 4; psses through the origin. 7 3 True Flse Determine whether the sttement is true or flse. Eplin our nswer. 7. A hperbol is the set of ll points in the plne tht re equidistnt from fied line nd fied point not on the line. 8. If n ellipse is not circle, then the foci of n ellipse lie on the mjor is of the ellipse. 9. If prbol hs eqution = 4p, where p is positive constnt, then the perpendiculr distnce from the prbol s focus to its directri is p. 3. The hperbol ( / ) = hs smptotes the lines =±/. 3. () As illustrted in the ccompning figure, prbolic rch spns rod 4 ft wide. How high is the rch if center section of the rod ft wide hs minimum clernce of ft? (b) How high would the center be if the rch were the upper hlf of n ellipse? ft ft ft 4 ft Figure E-3

334 746 Chpter / Prmetric nd Polr Curves; Conic Sections 3. () Find n eqution for the prbolic rch with bse b nd height h, shown in the ccompning figure. (b) Find the re under the rch. b, h (b, ) Figure E Show tht the verte is the closest point on prbol to the focus. [Suggestion: Introduce convenient coordinte sstem nd use Definition.4..] 34. As illustrted in the ccompning figure, suppose tht comet moves in prbolic orbit with the Sun t its focus nd tht the line from the Sun to the comet mkes n ngle of 6 with the is of the prbol when the comet is 4 million miles from the center of the Sun. Use the result in Eercise 33 to determine how close the comet will come to the center of the Sun. 35. For the prbolic reflector in the ccompning figure, how fr from the verte should the light source be plced to produce bem of prllel rs? 6 ft Figure E-34 Figure E () Show tht the right nd left brnches of the hperbol b = cn be represented prmetricll s = cosh t, = b sinh t ( <t<+ ) = cosh t, = b sinh t ( <t<+ ) (b) Use grphing utilit to generte both brnches of the hperbol = on the sme screen. 37. () Show tht the right nd left brnches of the hperbol b = cn be represented prmetricll s = sec t, = b tn t ( π/ <t<π/) = sec t, = b tn t ( π/ <t<π/) (b) Use grphing utilit to generte both brnches of the hperbol = on the sme screen. 38. Find n eqution of the prbol trced b point tht moves so tht its distnce from (, 4) is the sme s its distnce to the -is. ft 39. Find n eqution of the ellipse trced b point tht moves so tht the sum of its distnces to (4, ) nd (4, 5) is. 4. Find the eqution of the hperbol trced b point tht moves so tht the difference between its distnces to (, ) nd (, ) is. 4. Suppose tht the bse of solid is ellipticl with mjor is of length 9 nd minor is of length 4. Find the volume of the solid if the cross sections perpendiculr to the mjor is re squres (see the ccompning figure). 4. Suppose tht the bse of solid is ellipticl with mjor is of length 9 nd minor is of length 4. Find the volume of the solid if the cross sections perpendiculr to the minor is re equilterl tringles (see the ccompning figure). Figure E-4 Figure E Show tht n ellipse with semimjor is nd semiminor is b hs re A = πb. FOCUS ON CONCEPTS 44. Show tht if plne is not prllel to the is of right circulr clinder, then the intersection of the plne nd clinder is n ellipse (possibl circle). [Hint: Let θ be the ngle shown in the ccompning figure, introduce coordinte es s shown, nd epress nd in terms of nd.] u Figure E As illustrted in the ccompning figure on the net pge, crpenter needs to cut n ellipticl hole in sloped roof through which circulr vent pipe of dimeter D is to be inserted verticll. The crpenter wnts to drw the outline of the hole on the roof using pencil, two tcks, nd piece of string (s in Figure.4.3b). The center point of the ellipse is known, nd common sense suggests tht its mjor is must be perpendiculr to the drip line of the roof. The crpenter needs to determine the length L of the string nd the distnce T between tck nd the center point. The rchitect s plns show tht the pitch of the roof is p (pitch = rise over run; see the ccompning figure). Find T nd L in terms of D nd p. Source: This eercise is bsed on n rticle b Willim H. Enos, which ppered in the Mthemtics Techer, Feb. 99, p. 48.

335 .4 Conic Sections 747 Vent pipe Drip line Run Rise Figure E As illustrted in the ccompning figure, suppose tht two observers re sttioned t the points F (c, ) nd F ( c, ) in n -coordinte sstem. Suppose lso tht the sound of n eplosion in the -plne is herd b the F observer t seconds before it is herd b the F observer. Assuming tht the speed of sound is constnt v, show tht the eplosion occurred somewhere on the hperbol v t /4 c (v t /4) = C 5. () Sketch the solid generted b revolving R bout the -is, nd find its volume. (b) Sketch the solid generted b revolving R bout the -is, nd find its volume. As illustrted in the ccompning figure, the tnk of n oil truck is 8 ft long nd hs ellipticl cross sections tht re 6 ft wide nd 4 ft high. () Show tht the volume V of oil in the tnk (in cubic feet) when [ it is filled to depth of h feet is V = 7 4 sin h + (h ) ] 4h h + π (b) Use the numericl root-finding cpbilit of CAS to determine how mn inches from the bottom of dipstick the clibrtion mrks should be plced to indicte when the tnk is 4,, nd 3 4 full. 8 Dipstick 4 F ( c, ) F (c, ) Figure E As illustrted in the ccompning figure, suppose tht two trnsmitting sttions re positioned km prt t points F (5, ) nd F ( 5, ) on stright shoreline in n -coordinte sstem. Suppose lso tht ship is trveling prllel to the shoreline but km t se. Find the coordintes of the ship if the sttions trnsmit pulse simultneousl, but the pulse from sttion F is received b the ship microseconds sooner thn the pulse from sttion F. [Assume tht the pulses trvel t the speed of light (99,79,458 m/s).] km F ( 5, ) F (5, ) Figure E A nucler cooling tower is to hve height of h feet nd the shpe of the solid tht is generted b revolving the region R enclosed b the right brnch of the hperbol 5 5 = 34,5 nd the lines =, = h/, nd = h/ bout the -is. () Find the volume of the tower. (b) Find the lterl surfce re of the tower. 49. Let R be the region tht is bove the -is nd enclosed between the curve b = b nd the line = + b. Figure E-5 5. Prove: The line tngent to the prbol = 4p t the point (, ) is = p( + ). 5. Prove: The line tngent to the ellipse + b = t the point (, ) hs the eqution + b = 53. Prove: The line tngent to the hperbol b = t the point (, ) hs the eqution b = 54. Use the results in Eercises 5 nd 53 to show tht if n ellipse nd hperbol hve the sme foci, then t ech point of intersection their tngent lines re perpendiculr. 55. Find two vlues of k such tht the line + = k is tngent to the ellipse + 4 = 8. Find the points of tngenc. 56. Find the coordintes of ll points on the hperbol 4 = 4 where the two lines tht pss through the point nd the foci re perpendiculr. 57. A line tngent to the hperbol 4 = 36 intersects the -is t the point (, 4). Find the point(s) of tngenc. 58. Consider the second-degree eqution A + C + D + E + F = h 6 (cont.)

336 748 Chpter / Prmetric nd Polr Curves; Conic Sections where A nd C re not both. Show b completing the squre: () If AC >, then the eqution represents n ellipse, circle, point, or hs no grph. (b) If AC <, then the eqution represents hperbol or pir of intersecting lines. (c) If AC =, then the eqution represents prbol, pir of prllel lines, or hs no grph. 59. In ech prt, use the result in Eercise 58 to mke sttement bout the grph of the eqution, nd then check our conclusion b completing the squre nd identifing the grph. () 5 9 = (b) = (c) = (d) = (e) = (f ) = 6. Derive the eqution = 4p in Figure Derive the eqution ( /b ) + ( / ) = given in Figure Derive the eqution ( / ) ( /b ) = given in Figure Prove Theorem.4.4. [Hint: Choose coordinte es so tht the prbol hs the eqution = 4p. Show tht the tngent line t P(, ) intersects the -is t Q(, ) nd tht the tringle whose three vertices re t P,Q, nd the focus is isosceles.] 64. Given two intersecting lines, let L be the line with the lrger ngle of inclintion φ, nd let L be the line with the smller ngle of inclintion φ. We define the ngle θ between L nd L b θ = φ φ. (See the ccompning figure.) () Prove: If L nd L re not perpendiculr, then tn θ = m m + m m where L nd L hve slopes m nd m. (b) Prove Theorem.4.5. [Hint: Introduce coordintes so tht the eqution ( / ) + ( /b ) = describes the ellipse, nd use prt ().] (c) Prove Theorem.4.6. [Hint: Introduce coordintes so tht the eqution ( / ) ( /b ) = describes the hperbol, nd use prt ().] L u f L f Figure E Writing Suppose tht ou wnt to drw n ellipse tht hs given vlues for the lengths of the mjor nd minor es b using the method shown in Figure.4.3b. Assuming tht the es re drwn, eplin how compss cn be used to locte the positions for the tcks. 66. Writing List the forms for stndrd equtions of prbols, ellipses, nd hperbols, nd write summr of techniques for sketching conic sections from their stndrd equtions. QUICK CHECK ANSWERS.4. () n ellipse (b) hperbol (c) prbol. () = 4p (b) = 4p 3. () (b) + = ; b b + = (c) b 4. () (b) = ; b b = (c) + b ; b.5 ROTATION OF AXES; SECOND-DEGREE EQUATIONS In the preceding section we obtined equtions of conic sections with es prllel to the coordinte es. In this section we will stud the equtions of conics tht re tilted reltive to the coordinte es. This will led us to investigte rottions of coordinte es. QUADRATIC EQUATIONS IN AND We sw in Emples 8 to of the preceding section tht equtions of the form A + C + D + E + F = () cn represent conic sections. Eqution () is specil cse of the more generl eqution A + B + C + D + E + F = () which, if A, B, nd C re not ll zero, is clled qudrtic eqution in nd. It is usull the cse tht the grph of n second-degree eqution is conic section. If B =,

337 P(, ) (, ) Figure.5. 3 (, ).5 Rottion of Aes; Second-Degree Equtions 749 then () reduces to () nd the conic section hs its is or es prllel to the coordinte es. However, if B =, then () contins cross-product term B, nd the grph of the conic section represented b the eqution hs its is or es tilted reltive to the coordinte es. As n illustrtion, consider the ellipse with foci F (, ) nd F (, ) nd such tht the sum of the distnces from ech point P(,) on the ellipse to the foci is 6 units. Epressing this condition s n eqution, we obtin (Figure.5.) ( ) + ( ) + ( + ) + ( + ) = 6 Squring both sides, then isolting the remining rdicl, then squring gin ultimtel ields = 36 s the eqution of the ellipse. This is of form () with A = 8,B = 4,C = 5,D =, E =, nd F = 36. ROTATION OF AXES To stud conics tht re tilted reltive to the coordinte es it is frequentl helpful to rotte the coordinte es, so tht the rotted coordinte es re prllel to the es of the conic. Before we cn discuss the detils, we need to develop some ides bout rottion of coordinte es. In Figure.5. the es of n -coordinte sstem hve been rotted bout the origin through n ngle θ to produce new -coordinte sstem. As shown in the figure, ech point P in the plne hs coordintes (, ) s well s coordintes (, ). To see how the two re relted, let r be the distnce from the common origin to the point P, nd let α be the ngle shown in Figure.5.b. It follows tht = r cos(θ + α), = r sin(θ + α) (3) nd = r cos α, = r sin α (4) Using fmilir trigonometric identities, the reltionships in (3) cn be written s = r cos θ cos α r sin θ sin α = r sin θ cos α + r cos θ sin α nd on substituting (4) in these equtions we obtin the following reltionships clled the rottion equtions: = cos θ sin θ = sin θ + (5) cos θ ' P (, ) (', ') ' ' P ' ' r u u ' Figure.5. () (b) Emple Suppose tht the es of n -coordinte sstem re rotted through n ngle of θ = 45 to obtin n -coordinte sstem. Find the eqution of the curve + 6 = in -coordintes.

338 75 Chpter / Prmetric nd Polr Curves; Conic Sections ' ' Solution. Substituting sin θ = sin 45 = / nd cos θ = cos 45 = / in (5) ields the rottion equtions = nd = + Figure = Substituting these into the given eqution ields ( ) or or ( )( + ) + ( + ) 6 = = = which is the eqution of n ellipse (Figure.5.3). If the rottion equtions (5) re solved for nd in terms of nd, one obtins (Eercise 6): = cos θ + sin θ (6) = sin θ + cos θ Emple Find the new coordintes of the point (, 4) if the coordinte es re rotted through n ngle of θ = 3. Solution. Using the rottion equtions in (6) with =, = 4, cos θ = cos 3 = 3/, nd sin θ = sin 3 = /, we obtin = ( 3/) + 4(/) = 3 + = (/) + 4( 3/) = + 3 Thus, the new coordintes re ( 3 +, + 3 ). ELIMINATING THE CROSS-PRODUCT TERM In Emple we were ble to identif the curve + 6 = s n ellipse becuse the rottion of es eliminted the -term, thereb reducing the eqution to fmilir form. This occurred becuse the new -es were ligned with the es of the ellipse. The following theorem tells how to determine n pproprite rottion of es to eliminte the cross-product term of second-degree eqution in nd. It is lws possible to stisf (8) with n ngle θ in the intervl <θ<π/ We will lws choose θ in this w..5. theorem If the eqution A + B + C + D + E + F = (7) is such tht B =, nd if n -coordinte sstem is obtined b rotting the -es through n ngle θ stisfing cot θ = A C (8) B then, in -coordintes, Eqution (7) will hve the form A + C + D + E + F = proof Substituting (5) into (7) nd simplifing ields A + B + C + D + E + F =

339 .5 Rottion of Aes; Second-Degree Equtions 75 where A = A cos θ + B cos θ sin θ + C sin θ B = B(cos θ sin θ)+ (C A) sin θ cos θ C = A sin θ B sin θ cos θ + C cos θ D = D cos θ + E sin θ E = D sin θ + E cos θ F = F (9) (Verif.) To complete the proof we must show tht B = if or, equivlentl, cot θ = A C B cos θ sin θ = A C B However, b using the trigonometric double-ngle formuls, we cn rewrite B in the form () B = B cos θ (A C)sin θ Thus, B = ifθ stisfies (). Emple 3 Identif nd sketch the curve =. Solution. As first step, we will rotte the coordinte es to eliminte the cross-product term. Compring the given eqution to (7), we hve A =, B =, C = Thus, the desired ngle of rottion must stisf ' = ' cot θ = A C B = This condition cn be met b tking θ = π/orθ = π/4 = 45. Mking the substitutions cos θ = cos 45 = / nd sin θ = sin 45 = / in (5) ields = Figure.5.4 = nd = + Substituting these in the eqution = ields ( )( ) + = or = which is the eqution in the -coordinte sstem of n equilterl hperbol with vertices t (, ) nd (, ) in tht coordinte sstem (Figure.5.4). In problems where it is inconvenient to solve cot θ = A C B for θ, the vlues of sin θ nd cos θ needed for the rottion equtions cn be obtined b first clculting cos θ nd then computing sin θ nd cos θ from the identities sin θ = cos θ nd cos θ = + cos θ

340 75 Chpter / Prmetric nd Polr Curves; Conic Sections 4 7 Figure.5.5 ' Figure u (' ) + ' = 9 There is method for deducing the kind of curve represented b seconddegree eqution directl from the eqution itself without rotting coordinte es. For discussion of this topic, see the section on the discriminnt tht ppers in Web Appendi K. ' Emple 4 Solution. Identif nd sketch the curve = We hve A = 53,B = 9, nd C = 97, so cot θ = A C B = 56 9 = 7 4 Since θ is to be chosen in the rnge <θ<π/, this reltionship is represented b the tringle in Figure.5.5. From tht tringle we obtin cos θ = 7, which implies tht 5 + cos θ 7 5 cos θ = = = 3 5 cos θ sin θ = = = 4 5 Substituting these vlues in (5) ields the rottion equtions = nd = nd substituting these in turn in the given eqution ields 53 5 (3 4 ) 9 5 (3 4 )(4 + 3 ) (4 + 3 ) which simplifies to or 3 5 (3 4 ) 4 5 (4 + 3 ) = = = Completing the squre ields ( ) + = 9 which is the eqution in the -coordinte sstem of n ellipse with center (, ) in tht coordinte sstem nd semies = 3 nd b = (Figure.5.6). QUICK CHECK EXERCISES.5 (See pge 754 for nswers.). Suppose tht n -coordinte sstem is rotted θ rdins to produce new -coordinte sstem. () nd m be obtined from,, nd θ using the rottion equtions = nd =. (b) nd m be obtined from,, nd θ using the equtions = nd =.. If the eqution A + B + C + D + E + F = is such tht B =, then the -term in this eqution cn be eliminted b rottion of es through n ngle θ stisfing cot θ =. 3. In ech prt, determine rottion ngle θ tht will eliminte the -term. () = (b) = (c) = 4. Epress + + = inthe -coordinte sstem obtined b rotting the -coordinte sstem through the ngle θ = π/4.

341 .5 Rottion of Aes; Second-Degree Equtions 753 EXERCISE SET.5. Let n -coordinte sstem be obtined b rotting n -coordinte sstem through n ngle of θ = 6. () Find the -coordintes of the point whose -coordintes re (, 6). (b) Find n eqution of the curve 3 + = 6 in -coordintes. (c) Sketch the curve in prt (b), showing both -es nd -es.. Let n -coordinte sstem be obtined b rotting n -coordinte sstem through n ngle of θ = 3. () Find the -coordintes of the point whose -coordintes re (, 3). (b) Find n eqution of the curve + 3 = 3in -coordintes. (c) Sketch the curve in prt (b), showing both -es nd -es. 3 Rotte the coordinte es to remove the -term. Then identif the tpe of conic nd sketch its grph. 3. = = = = = = = = = = 3. Let n -coordinte sstem be obtined b rotting n coordinte sstem through n ngle of 45. Use (6) to find n eqution of the curve 3 + = 6in-coordintes. 4. Let n -coordinte sstem be obtined b rotting n -coordinte sstem through n ngle of 3. Use (5) to find n eqution in -coordintes of the curve =. FOCUS ON CONCEPTS 5. Let n -coordinte sstem be obtined b rotting n -coordinte sstem through n ngle θ. Prove: For ever vlue of θ, the eqution + = r becomes the eqution + = r. Give geometric eplntion. 6. Derive (6) b solving the rottion equtions in (5) for nd in terms of nd. 7. Let n -coordinte sstem be obtined b rotting n -coordinte sstem through n ngle θ. Eplin how to find the -coordintes of point whose - coordintes re known. 8. Let n -coordinte sstem be obtined b rotting n -coordinte sstem through n ngle θ. Eplin how to find the -eqution of line whose -eqution is known. 9 Show tht the grph of the given eqution is prbol. Find its verte, focus, nd directri = = = = 3 6 Show tht the grph of the given eqution is n ellipse. Find its foci, vertices, nd the ends of its minor is = = = = 7 3 Show tht the grph of the given eqution is hperbol. Find its foci, vertices, nd smptotes = = = = 3. Show tht the grph of the eqution + = is portion of prbol. [Hint: First rtionlize the eqution nd then perform rottion of es.] FOCUS ON CONCEPTS 3. Derive the epression for B in (9). 33. Use (9) to prove tht B 4AC = B 4A C for ll vlues of θ. 34. Use (9) to prove tht A + C = A + C for ll vlues of θ. 35. Prove: If A = C in (7), then the cross-product term cn be eliminted b rotting through Prove: If B =, then the grph of + B + F = is hperbol if F = nd two intersecting lines if F =.

342 754 Chpter / Prmetric nd Polr Curves; Conic Sections QUICK CHECK ANSWERS.5. () cos θ sin θ; sin θ + cos θ (b) cos θ + sin θ; sin θ + cos θ = A C B 3. () π 4 (b) π 3 (c) π 6.6 CONIC SECTIONS IN POLAR COORDINATES It will be shown lter in the tet tht if n object moves in grvittionl field tht is directed towrd fied point (such s the center of the Sun), then the pth of tht object must be conic section with the fied point t focus. For emple, plnets in our solr sstem move long ellipticl pths with the Sun t focus, nd the comets move long prbolic, ellipticl, or hperbolic pths with the Sun t focus, depending on the conditions under which the were born. For pplictions of this tpe it is usull desirble to epress the equtions of the conic sections in polr coordintes with the pole t focus. In this section we will show how to do this. THE FOCUS DIRECTRIX CHARACTERIZATION OF CONICS To obtin polr equtions for the conic sections we will need the following theorem. It is n unfortunte historicl ccident tht the letter e is used for the bse of the nturl logrithm s well s for the eccentricit of conic sections. However, s prcticl mtter the pproprite interprettion will usull be cler from the contet in which the letter is used..6. theorem (Focus Directri Propert of Conics) Suppose tht point P moves in the plne determined b fied point (clled the focus) nd fied line (clled the directri), where the focus does not lie on the directri. If the point moves in such w tht its distnce to the focus divided b its distnce to the directri is some constnt e (clled the eccentricit), then the curve trced b the point is conic section. Moreover, the conic is () prbol if e = (b) n ellipse if <e< (c) hperbol if e>. We will not give forml proof of this theorem; rther, we will use the specific cses in Figure.6. to illustrte the bsic ides. For the prbol, we will tke the directri to be = p, s usul; nd for the ellipse nd the hperbol we will tke the directri to be = /c. We wnt to show in ll three cses tht if P is point on the grph, F is the focus, nd D is the directri, then the rtio PF/PD is some constnt e, where e = for the prbol, <e< for the ellipse, nd e> for the hperbol. We will give the rguments for the prbol nd ellipse nd leve the rgument for the hperbol s n eercise. D P(, ) P(, ) D D P(, ) F(p, ) F(c, ) F(c, ) = p = /c e = < e < e > = /c Figure.6.

343 .6 Conic Sections in Polr Coordintes 755 For the prbol, the distnce PF to the focus is equl to the distnce PD to the directri, so tht PF/PD =, which is wht we wnted to show. For the ellipse, we rewrite Eqution (8) of Section.4 s ( c) + = c = c ( ) c But the epression on the left side is the distnce PF, nd the epression in the prentheses on the right side is the distnce PD, so we hve shown tht PF = c PD Thus, PF/PD is constnt, nd the eccentricit is e = c () If we rule out the degenerte cse where = orc =, then it follows from Formul (7) of Section.4 tht <c<,so <e<, which is wht we wnted to show. We will leve it s n eercise to show tht the eccentricit of the hperbol in Figure.6. is lso given b Formul (), but in this cse it follows from Formul () of Section.4 tht c>,so e>. ECCENTRICITY OF AN ELLIPSE AS A MEASURE OF FLATNESS The eccentricit of n ellipse cn be viewed s mesure of its fltness s e pproches the ellipses become more nd more circulr, nd s e pproches the become more nd more flt (Figure.6.). Tble.6. shows the orbitl eccentricities of vrious celestil objects. Note tht most of the plnets ctull hve firl circulr orbits. e = F e =.5 Ellipses with common focus nd equl semimjor es Figure.6. e =.8 e =.9 Tble.6. celestil bod Mercur Venus Erth Mrs Jupiter Sturn Urnus Neptune Pluto Hlle's comet eccentricit F Pole r u r cos u Figure.6.3 P(r, u) d D Directri POLAR EQUATIONS OF CONICS Our net objective is to derive polr equtions for the conic sections from their focus directri chrcteriztions. We will ssume tht the focus is t the pole nd the directri is either prllel or perpendiculr to the polr is. If the directri is prllel to the polr is, then it cn be bove or below the pole; nd if the directri is perpendiculr to the polr is, then it cn be to the left or right of the pole. Thus, there re four cses to consider. We will derive the formuls for the cse in which the directri is perpendiculr to the polr is nd to the right of the pole. As illustrted in Figure.6.3, let us ssume tht the directri is perpendiculr to the polr is nd d units to the right of the pole, where the constnt d is known. If P is point

344 756 Chpter / Prmetric nd Polr Curves; Conic Sections on the conic nd if the eccentricit of the conic is e, then it follows from Theorem.6. tht PF/PD = e or, equivlentl, tht PF = epd () However, it is evident from Figure.6.3 tht PF = r nd PD = d r cos θ.thus, () cn be written s r = e(d r cos θ) which cn be solved for r nd epressed s r = ed + e cos θ (verif). Observe tht this single polr eqution cn represent prbol, n ellipse, or hperbol, depending on the vlue of e. In contrst, the rectngulr equtions for these conics ll hve different forms. The derivtions in the other three cses re similr..6. theorem If conic section with eccentricit e is positioned in polr coordinte sstem so tht its focus is t the pole nd the corresponding directri is d units from the pole nd is either prllel or perpendiculr to the polr is, then the eqution of the conic hs one of four possible forms, depending on its orienttion: Directri Directri Focus Focus r = ed + e cos θ Directri right of pole ed r = e cos θ Directri left of pole (3 4) Focus Directri ed r = + e sin θ Directri bove pole Focus Directri ed r = e sin θ Directri below pole (5 6) SKETCHING CONICS IN POLAR COORDINATES Precise grphs of conic sections in polr coordintes cn be generted with grphing utilities. However, it is often useful to be ble to mke quick sketches of these grphs tht show their orienttions nd give some sense of their dimensions. The orienttion of conic reltive to the polr is cn be deduced b mtching its eqution with one of the four forms in Theorem.6.. The ke dimensions of prbol re determined b the constnt p (Figure.4.5) nd those of ellipses nd hperbols b the constnts,b, nd c (Figures.4. nd.4.). Thus, we need to show how these constnts cn be obtined from the polr equtions.

345 Emple Sketch the grph of r =.6 Conic Sections in Polr Coordintes 757 cos θ in polr coordintes. Solution. The eqution is n ect mtch to (4) with d = nd e =. Thus, the grph is prbol with the focus t the pole nd the directri units to the left of the pole. This tells us tht the prbol opens to the right long the polr is nd p =. Thus, the prbol looks roughl like tht sketched in Figure.6.4. Figure.6.4 Figure.6.5 Rough sketch b r In words, Formul (8) sttes tht is the rithmetic verge (lso clled the rithmetic men) of r nd r, nd Formul () sttes tht b is the geometric men of r nd r. Figure.6.6 Rough sketch b c c r All of the importnt geometric informtion bout n ellipse cn be obtined from the vlues of,b, nd c in Figure.6.5. One w to find these vlues from the polr eqution of n ellipse is bsed on finding the distnces from the focus to the vertices. As shown in the figure, let r be the distnce from the focus to the closest verte nd r the distnce to the frthest verte. Thus, from which it follows tht Moreover, it lso follows from (7) tht Thus, r = c nd r = + c (7) = (r + r ) c = (r r ) (8 9) r r = c = b b = r r () Emple Find the constnts, b, nd c for the ellipse r = 6 + cos θ. Solution. This eqution does not mtch n of the forms in Theorem.6. becuse the ll require constnt term of in the denomintor. However, we cn put the eqution into one of these forms b dividing the numertor nd denomintor b to obtin 3 r = + cos θ This is n ect mtch to (3) with d = 6 nd e =, so the grph is n ellipse with the directri 6 units to the right of the pole. The distnce r from the focus to the closest verte cn be obtined b setting θ = in this eqution, nd the distnce r to the frthest verte cn be obtined b setting θ = π. This ields 3 r = + cos = 3 3 =, r 3 = + cos π = 3 = 6 Thus, from Formuls (8), (), nd (9), respectivel, we obtin = (r + r ) = 4, b = r r = 3, c = (r r ) = The ellipse looks roughl like tht sketched in Figure.6.6. Figure.6.7 r r All of the importnt informtion bout hperbol cn be obtined from the vlues of,b, nd c in Figure.6.7. As with the ellipse, one w to find these vlues from the polr eqution of hperbol is bsed on finding the distnces from the focus to the vertices. As

346 758 Chpter / Prmetric nd Polr Curves; Conic Sections shown in the figure, let r be the distnce from the focus to the closest verte nd r the distnce to the frthest verte. Thus, from which it follows tht r = c nd r = c + () = (r r ) c = (r + r ) ( 3) In words, Formul (3) sttes tht c is the rithmetic men of r nd r, nd Formul (4) sttes tht b is the geometric men of r nd r. Moreover, it lso follows from () tht r r = c = b from which it follows tht b = r r (4) Emple 3 Sketch the grph of r = + sin θ in polr coordintes. Solution. This eqution is n ect mtch to (5) with d = nd e =. Thus, the grph is hperbol with its directri unit bove the pole. However, it is not so strightforwrd to compute the vlues of r nd r, since hperbols in polr coordintes re generted in strnge w s θ vries from to π. This cn be seen from Figure.6.8, which is the grph of the given eqution in rectngulr θr-coordintes. It follows from this grph tht the corresponding polr grph is generted in pieces (see Figure.6.8b): As θ vries over the intervl θ<7π/6, the vlue of r is positive nd vries from downto/3 nd then to +, which genertes prt of the lower brnch. As θ vries over the intervl 7π/6 <θ 3π/, the vlue of r is negtive nd vries from to, which genertes the right prt of the upper brnch. As θ vries over the intervl 3π/ θ<π/6, the vlue of r is negtive nd vries from to, which genertes the left prt of the upper brnch. As θ vries over the intervl π/6 <θ π, the vlue of r is positive nd vries from + to, which fills in the missing piece of the lower right brnch. 3 r 6 e i m o u r = + sin u Rough sketch Rough sketch () (b) (c) Figure.6.8 To obtin rough sketch of hperbol, it is generll sufficient to locte the center, the smptotes, nd the points where θ =, θ = π/, θ = π, nd θ = 3π/. It is now cler tht we cn obtin r b setting θ = π/ nd r b setting θ = 3π/. Keeping in mind tht r nd r re positive, this ields r = + sin(π/) = 3, r = + sin(3π/) = =

347 Thus, from Formuls (), (4), nd (3), respectivel, we obtin.6 Conic Sections in Polr Coordintes 759 = (r r ) = 3, b = r r = 3 3, c = (r + r ) = 4 3 Thus, the hperbol looks roughl like tht sketched in Figure.6.8c. APPLICATIONS IN ASTRONOMY In 69 Johnnes Kepler published book known s Astronomi Nov (or sometimes Commentries on the Motions of Mrs) in which he succeeded in distilling thousnds of ers of observtionl stronom into three beutiful lws of plnetr motion (Figure.6.9). Sun Equl res re swept out in equl times, nd the squre of the period T is proportionl to kepler s lws First lw (Lw of Orbits). Ech plnet moves in n ellipticl orbit with the Sun t focus. Second lw (Lw of Ares). The rdil line from the center of the Sun to the center of plnet sweeps out equl res in equl times. Third lw (Lw of Periods). The squre of plnet s period (the time it tkes the plnet to complete one orbit bout the Sun) is proportionl to the cube of the semimjor is of its orbit. Figure.6.9 Apogee Figure.6. Perigee Kepler s lws, lthough stted for plnetr motion round the Sun, ppl to ll orbiting celestil bodies tht re subjected to single centrl grvittionl force rtificil stellites subjected onl to the centrl force of Erth s grvit nd moons subjected onl to the centrl grvittionl force of plnet, for emple. Lter in the tet we will derive Kepler s lws from bsic principles, but for now we will show how the cn be used in bsic stronomicl computtions. In n ellipticl orbit, the closest point to the focus is clled the perigee nd the frthest point the pogee (Figure.6.). The distnces from the focus to the perigee nd pogee Johnnes Kepler (57 63) Germn stronomer nd phsicist. Kepler, whose work provided our contemporr view of plnetr motion, led fscinting but ill-strred life. His lcoholic fther mde him work in fmil-owned tvern s child, lter withdrwing him from elementr school nd hiring him out s field lborer, where the bo contrcted smllpo, permnentl crippling his hnds nd impiring his eesight. In lter ers, Kepler s first wife nd severl children died, his mother ws ccused of witchcrft, nd being Protestnt he ws often subjected to persecution b Ctholic uthorities. He ws often impoverished, eking out living s n strologer nd prognostictor. Looking bck on his unhpp childhood, Kepler described his fther s criminll inclined nd qurrelsome nd his mother s grrulous nd bd-tempered. However, it ws his mother who left n indelible mrk on the sier-old Kepler b showing him the comet of 577; nd in lter life he personll prepred her defense ginst the witchcrft chrges. Kepler becme cquinted with the work of Copernicus s student t the Universit of Tübingen, where he received his mster s degree in 59. He continued on s theologicl student, but t the urging of the universit officils he bndoned his clericl studies nd ccepted position s mthemticin nd techer in Grz, Austri. However, he ws epelled from the cit when it cme under Ctholic control, nd in 6 he finll moved on to Prgue, where he becme n ssistnt t the observtor of the fmous Dnish stronomer Tcho Brhe. Brhe ws brillint nd meticulous stronomicl observer who mssed the most ccurte stronomicl dt known t tht time; nd when Brhe died in 6 Kepler inherited the tresure-trove of dt. After eight ers of intense lbor, Kepler deciphered the underling principles buried in the dt nd in 69 published his monumentl work, Astronomi Nov, in which he stted his first two lws of plnetr motion. Commenting on his discover of ellipticl orbits, Kepler wrote, I ws lmost driven to mdness in considering nd clculting this mtter. I could not find out wh the plnet would rther go on n ellipticl orbit (rther thn circle). Oh ridiculous me! It ultimtel remined for Isc Newton to discover the lws of grvittion tht eplined the reson for ellipticl orbits. [Imge:

348 76 Chpter / Prmetric nd Polr Curves; Conic Sections Center Figure.6. e Directri Focus e re clled the perigee distnce nd pogee distnce, respectivel. For orbits round the Sun, it is more common to use the terms perihelion nd phelion, rther thn perigee nd pogee, nd to mesure time in Erth ers nd distnces in stronomicl units (AU), where AU is the semimjor is of the Erth s orbit (pproimtel 5 6 km or mi). With this choice of units, the constnt of proportionlit in Kepler s third lw is, since = AU produces period of T = Erth er. In this cse Kepler s third lw cn be epressed s T = 3/ (5) Shpes of ellipticl orbits re often specified b giving the eccentricit e nd the semimjor is, so it is useful to epress the polr equtions of n ellipse in terms of these constnts. Figure.6., which cn be obtined from the ellipse in Figure.6. nd the reltionship c = e, implies tht the distnce d between the focus nd the directri is d = e c = e e = ( e ) (6) e from which it follows tht ed = ( e ). Thus, depending on the orienttion of the ellipse, the formuls in Theorem.6. cn be epressed in terms of nd e s r = ( e ) ± e cos θ +: Directri right of pole : Directri left of pole r = ( e ) ± e sin θ +: Directri bove pole : Directri below pole (7 8) Moreover, it is evident from Figure.6. tht the distnces from the focus to the closest nd frthest vertices cn be epressed in terms of nd e s r = e = ( e) nd r = + e = ( + e) (9 ) Hlle's comet c/ Emple 4 Hlle s comet (lst seen in 986) hs n eccentricit of.97 nd semimjor is of = 8.AU. Figure.6. () Find the eqution of its orbit in the polr coordinte sstem shown in Figure.6.. (b) Find the period of its orbit. (c) Find its perihelion nd phelion distnces. Solution (). From (7), the polr eqution of the orbit hs the form r = ( e ) + e cos θ But ( e ) = 8.[ (.97) ].7. Thus, the eqution of the orbit is.7 r = +.97 cos θ Science Photo Librr/Photo Reserchers Hlle s comet photogrphed April, 9 in Peru. Solution (b). From (5), with = 8., the period of the orbit is T = (8.) 3/ 77 ers Solution (c). Since the perihelion nd phelion distnces re the distnces to the closest nd frthest vertices, respectivel, it follows from (9) nd () tht r = e = ( e) = 8.(.97).543 AU r = + e = ( + e) = 8.( +.97) 35.7AU

349 .6 Conic Sections in Polr Coordintes 76 or since AU 5 6 km, the perihelion nd phelion distnces in kilometers re r = 8.(.97)(5 6 ) 8,5, km r = 8.( +.97)(5 6 ) 5,35,, km Figure.6.3 [Imge: NASA] Minimum distnce Mimum distnce Emple 5 AnApollo lunr lnder orbits the Moon in n elliptic orbit with eccentricit e =. nd semimjor is = 5 km. Assuming the Moon to be sphere of rdius 74 km, find the minimum nd mimum heights of the lnder bove the lunr surfce (Figure.6.3). Solution. If we let r nd r denote the minimum nd mimum distnces from the center of the Moon, then the minimum nd mimum distnces from the surfce of the Moon will be d min = r 74 d m = r 74 or from Formuls (9) nd () d min = r 74 = ( e) 74 = 5(.88) 74 = 33. km d m = r 74 = ( + e) 74 = 5(.) 74 = 56.8 km QUICK CHECK EXERCISES.6 (See pge 763 for nswers.). In ech prt, nme the conic section described. () The set of points whose distnce to the point (, 3) is hlf the distnce to the line + = is. (b) The set of points whose distnce to the point (, 3) is equl to the distnce to the line + = is. (c) The set of points whose distnce to the point (, 3) is twice the distnce to the line + = is.. In ech prt: (i) Identif the polr grph s prbol, n ellipse, or hperbol; (ii) stte whether the directri is bove, below, to the left, or to the right of the pole; nd (iii) find the distnce from the pole to the directri. () r = (b) r = 4 + cos θ 4 cos θ 4 (c) r = (d) r = sin θ sin θ 3. If the distnce from verte of n ellipse to the nerest focus is r, nd if the distnce from tht verte to the frthest focus is r, then the semimjor is is = nd the semiminor is is b =. 4. If the distnce from verte of hperbol to the nerest focus is r, nd if the distnce from tht verte to the frthest focus is r, then the semifocl is is = nd the semiconjugte is is b =. EXERCISE SET.6 Grphing Utilit Find the eccentricit nd the distnce from the pole to the directri, nd sketch the grph in polr coordintes () r = (b) r = cos θ + sin θ 4 5. () r = (b) r = + 3 cos θ sin θ 3 4 Use Formuls (3) (6) to identif the tpe of conic nd its orienttion. Check our nswer b generting the grph with grphing utilit. 3. () r = 8 sin θ (b) r = sin θ 4. () r = 4 3 sin θ (b) r = 4 + cos θ 5 6 Find polr eqution for the conic tht hs its focus t the pole nd stisfies the stted conditions. Points re in polr coordintes nd directrices in rectngulr coordintes for simplicit. (In some cses there m be more thn one conic tht stisfies the conditions.) 5. () Ellipse; e = 3 ; directri =. 4 (b) Prbol; directri =. (c) Hperbol; e = 4 ; directri = 3. 3