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Calculus III

by: August Feeney

Calculus III MTH 253

August Feeney
GPA 3.98


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This 21 page Class Notes was uploaded by August Feeney on Monday October 19, 2015. The Class Notes belongs to MTH 253 at Portland Community College taught by Staff in Fall. Since its upload, it has received 60 views. For similar materials see /class/224649/mth-253-portland-community-college in Math at Portland Community College.

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Date Created: 10/19/15
Mr Simonds MTH 253 introduction to sequences and series The main Topics considered in MTH 253 revolve around The Two maThemaTicaI objecTs known as sequences and series IT is unforTunaTe ThaT These words are so similar in appearance and sound because while The Two objecTs are relaTed They are completely different Types of objects A sequence is an ordered set containing a neverending list of numbers A series is either a single number or is said to diverge A good way of Thinking abouT mosT sequences is ThaT They are never ending isTs of numbers generaTed by inserTing The posiTive inTegers inTo some maThemaTicaI formula illk k Find The firsT few Terms of The sequence defined by ak 3 Find a formula for The sequence 7 oolxl Slw 13 5 24 6 NoTice ThaT The Terms of The firsT sequence geT closer and closer To 3 as you move Through The isT where as The Terms of The second sequence bounce back and forTh beTween nearly 1 and nearly 71 This gives us our firsT peek aT The difference beTween convergenT sequences and divergenT sequences Definition If lim an L we say ThaT The sequence an converges To L Ham If lim an does noT exisT we say ThaT The sequence an diverges ThaT is a divergenT new sequence is a sequence ThaT does noT converge Page 1 of 6 Mr Simonds MTH 253 introduction to sequences and series LeT39s deTermine which of The following sequences are convergenT and which are divergenT If The sequence converges eT39s sTaTe The number To which The sequence converges 3k1 0k 4172 Dominant Term Analysis 7 3739k k720H 4k Dominant Term Analysis lk8 5k9 16k1 k WM Dominant Term Analysis Page 2 of 6 Definition Mr Simonds MTH 253 r introduction 0 Sequences and Series A recursive seguenc is a sequence where The value of The firsT Term and possibly a fewTerms following ThaT Term are explicile sTaTed and The value of subsequenT Terms are defined by a formula based upon The values of The Terms ThaT precede Them LeT39s wriTe ouT The firsT few Terms of The sequence ak 3 if k 1 a 17 M if k gt1 To whaT value does This sequence converge The Fibonacci sequence is The recursive sequence 0 ifk0 Fk 1 ifk1 Fk72FtH if kgt1 LeT39s lisT The firsT few Terms and illusTraTe a Fibonacci spiral wiTh The drawing on The 39g Page 3 of6 Mr Simonds MTH 253 introduction to sequences and series De nitions A series is The sum of The Terms of a sequence The nth parfial sum of a series is The sum of The firsT n Terms of The sequence being summed ThaT is if The sequence being summed is ak The nth parfial sum is Sn Z ak assuming k1 ThaT k sTarTs aT 1 Find The firsT five parTial sums of The series 2 k1 NoTice ThaT The sequence of parTial sums is recursive in naTure if noT in facT For The series 2 ak we have k 1 Sla1 SkSk1akifkgt1 1 Find The firsT five Terms in The sequence of EarTial sums for The series 2 To whaT number k1 does The sequence of parTial sums appear To limiT Page 4 of 6 Mr Simonds MTH 253 introduction to sequences and series Definitions A series is said To converge if iTs sequence of parTial sums is convergenT and The series is said To diverge if iTs sequence of parTial sums diverges If The sequence of parTial sums converges To The number S we say ThaT The sum of The series is S ThaT Is we say ThaT I ak ggquk S Consider The series 2 for The resT of This documenT k L k26k8 LisT The firsT 4 parTial sums 6 DeTermine The parTial fracTion decomposiTion of 2 k 6k8 Page 5 of 6 Mr Simonds MTH 253 introduction to sequences and series Explicile wriTe ouT 54 using ak Do noT reduce any of The fracTions Combine The 3 3 m 7 W39 fracTions ThaT have The same denominaTors buT do noT acTuay add or subTracT any of The oTher fracTions WriTe ouT The resulT RepeaT The lasT direcTions for SS WhaT39s a good guess for The value of 100 Verify The value on your calculaTor 26 kAE I Ek I B z k II 133 M l Mun ml39rn rimr nnn WhaT39s a good guess for The value of The series Page 6 of 6 Vectors lines and planes a focus on the basics da Key 1 2 Figure 1 Figure 2 Figure 3 Figure 4 3 4 Figure 5 Figure 6 Figure 7 1726 17755 17 6 7311 a750 94077 uv7577 17867ltO4L77810 1786 77473 av4 3 Page 1 of6 Vectors lines and planes a focus on the basics da Key 5 Use a proTracTor or your eyeball To esTimaTe The smallesT angle formed beTween The vecTors 171 and 7 a drawnon figures 8 6 Figure 8 Fiqure 9 Figure 10 1726 9915 a750 94077 86 v271 a J l COSquot i cos l cosquot A MR 5397 106 m 634 90 m 634 5 ltl lt5 8 cos l 5 lt5 lt5 w The resuITs maTch The angles originally drawn in figures 8 and 9 because The vecTors had already been drawn TailToTail To geT a maTch in Figure 10 we need To also draw The vecTors TailToTail NoTe ThaT The original angle is 180B 7 8 1166D We can apply The righT hand rLIIe sTraighT up in figures 11 and 12 In Figure 11 x 7 poinTs along The posiTive xaxis as iT does in Figure 12 To apply The righT hand rule in Figure 13 we need To redraw The vecTors TailToTail AfTer doing so a correcT applicaTion of The righT hand rLIIe reveals ThaT 171x 7 poinTs along The negaTive xaxis Page 2 of6 Vectors lines and planes a focus on the basics da Key 8 Figure 11 17026 7O755 axv4ooo Figure 12 ao75o 940077 ax lt35oo Figure 13 17lt086 a 0271 gtlt7lt72000 9 comp1 7 gt0 14 compI 7 lt 0 comp 7 lt 0 pm 10 Figure 14 17 26 7 755 compI 7 proJu 7 comp 503 0 proquotI 7 comp Vi 6 I 551797 5 Figure 15 compa 6 lul a75o 7lt077 a 7 713 26 Figure 16 213 l l m proj 7 comp 7 a51 77273comp 7 Figure 17 1705gt 7075comp 772575 proli compi775lt05075 Page 3 of6 Vectors lines and planes a focus on the basics da Key 11 u 6 x0 gt 4y76226 and 75yz713 ThepointAis 0273 IE i u i isms bl p u 1 u suiuoiuh x r XI 75 bllzlz39lX sq 2 y0 2 5225 and 2xz713 Thepointsis 4075 474724 d r q 1 P1 and i7 1 P2 r i1 xr u 445713713 Both V and E are multiples of 211 Vu Neatoi One set of equations x2t y2t 273 2 L42L763t 26 22 L752L3L 13 mm mm MD Mm rm mo Chec 2x75yz 1 1 Parametric equations for the line are x 9 31 y 7 7111 z 57 31 Substituting 1 1 into the plane equation gives us 29 31 7 7111 7 3 7 t 5 2 t 5 Substituting back into the parametric equations gives us the point 247487 2 24 e 7748 11 Checkline ii lvsdd Check piane 224 748 7 372 48 e 48 5 5 J The point of intersection is 247487 2 Page 4 of 6 Vectors lines and planes a focus on the basics da Key 13 Begin by idenTifying direcTion vecTors for each line and normal vecTors for each plane gi 3102 6 PJ 513 P5J 5 5 29 Pi 15 39 P7J 762 10 L1x714 12 L H 631 5 L H 6311 2 9 The 2 parallel planes musT have parallel normal vecTors This single facT narrows The field down To P3 and P6 which are The only 2 planes whose sTaTed normal vecTors are mulTipIes of one anoTher We really should verify ThaT P3 and P6 aren39T acTuaIIy The same plane Seeing as 070 lies on P3 buT noT on P6 They cerTainIy are noT The same plane Hence P3 and P6 are The parallel planes b The 2 perpendicular planes musT have perpendicular normal vecTors Looking for zero inner producTs we quickly esTainsh ThaT The perpendicular planes are P4 and P5 c A line and plane are perpendicular when The line39s direcTion vecTor is paraleTo The plane39s normal vecTor Looking for mulTipIes we quickly esTainsh ThaT P7 J L2 d and e If eiTher L P or L C P The line39s direcTion vecTor and The plane39s normal vecTor are perpendicular Looking for zero dinner producTs The 2 pairs up for consideraTion are L1 wiTh P5 and L3 wiTh P4 302 is on boTh L1 and P5 so L1 musT ie enTirer on P5 070 is on P4 buT noT L3 so L3 P4 a In The configuraTion below L7 W and L7 XW X W a lie on The page and L X W poinTs sTraighT ouT from The page This gives us X W X W X if X V TT v71 ParT b a 40 3 W 13 2 and a X W 9512 47 X W X W X a X w 463022 gtlt 9512 250750 500 250 Page 5 of 6 Vectors lines and planes 5 focus on the basics da Key sin where a is the 15 This entire problem is based upon the fact that Wm smaiiest angle formed when v and W are drawn taiitotaii One thing that is very relevant is that iz The other thing that is very relevant is that sin 6 0 a1 as 60 gt 90 iz2 x113 l 0 is the smallest number i 112SOIZ gtlt113gt11 gtlt112 iz Xizg is the largest number since i12 Tl i1 and i13 5 the largest the angle between the two vectors the greater Since iz izzli14 the magnitude of the cross product hence x112 is the largest number and iz2 mid is 939 the smallest number Page a uf a Mr Simonds MTH 253 Differential Equations Group Work Exercises 1 lquot 0 F 1 DeTermine which one of The slope fields on page 3 of The differenTial equaTions noTes dy corresponds To The equaTion 7 xy Then skeTch onTo The appropriaTe figure The soluTion curve ThaT passes Through The poinT 073 Finally find The general soluTion To 7 xy and deTermine ThaT The equaTion for The specific soluTion you skeTched is y Solve The iniTial value problem yl l1lZ 7l7ly y0 0 and show ThaT The soluTion can 71 be wriTTen in The form y NoTe To solve This problem you have no choice buT To dy assume ThaT The incompleTe symbol yquot39 is a subsTiTuTe for The meaningful expression l 12 e Z Solve The iniTial value problem y 7 y secx 0 y7 7 e Make sure ThaT you sTaTe your soluTion in The form y IceX E Coli likes To divide specifically when placed inTo a welcoming hosT each E Coli cell divides inTo 2 cells every 20 minuTes Assuming ThaT all cells divide on schedule and ThaT no cells die This means ThaT when placed inTo a welcoming hosT an E Coli populaTion doubles every 20 minuTes If we leT E be The number of E Coli cells presenT in Gomer39s Tummy 1 hours afTer he ingesTs 1500 E Coli cells and we assume ThaT Gomer39s Tummy is a welcoming hosT Then a dE formula for E can be found by solving The differenTial equaTion EkE Find The formula for E and use iT To deTermine how long iT Takes for Gomer39s 1500 cell sample To grow inTo a 1000000 cell sample Yummy cookies were pulled from a 175 C oven and were lefT To cool in a 22 C room In The firsT 5 minuTes one of The cookies cooled from 175 C To 99 C Once removed from The oven how much Time To The nearesT second did iT Take for This cookie To cool To wiThin 5 of The room TemperaTure Find The specific soluTion for The curve in Figure 11 Figure 11 y y 3d 4 3d 4 x 7xx y y y Page 1 of1 introduction to telescoping and geometric series Telescoging Series examples 4 Find The sum for The telesco in series P g k k2 4k 3 DayIA ah LL i 3 2 3 E 1 s L 3 3 39 1 1 39L I 1 Si 4 3quot Lt h39 L 1 39L L J 13 3 1 1 8 oleL39AZ L39OA 4 quot Page 1 of 8 1 Introduction to telescoping and geometric series L q 1 39 Kvl 3 9 z Use great bx I39llduc oI to prove the partial sum formula from page 1 a kll K 1 Show that the formula is valid at 1 2 Assume that the formula is valid foriy and show that it follows from this that the formula is also valid forlk 1 n 7 Z i d E 5 l5 HCMUI SI 3 3 391 1 r 39 rnl B c rh lk39 S 6quot E Avl uv 4639 39l39 V g 1 1 L sz 9 A quotquot 39 g L J an 3 S S 41 Q L AI 7 L 391 C 439 3935 n 39 7 13 quotquotquot39 1 00 1 39L 7 39L 33 T Au m L Mn3 K E I E 1 3 3 HHL quot Quin3 amp Esmb39ish The WWWon page 1 by hand LI A F L f 1Lr 4amp5 kw Kid J kr kfuuki J A 3 V 4 I klnlb ld 39 i Akin3 mm H KfeJ k0 q 39 3A 4393 Lf 2 L KLd 9 2 limo3 KN KP n g it 115 J n 3 8 1 1 Page 2 of 8 introduction to telescoping and geometric series quot 1 Find The sum for The Telescoping series 2 ln 1 k1 bug at M I t s 14 WK Mlt S al JAM new 39 xa LAisz A1L2L3 V H 1 Sl l L j xitl 53 514 Ms Nwi m 5 0 5 quot u It n b 2 AUT IA39 2 14 440 J Page 3 of 8 introduction to telescoping and geometric series Find The sum for The Telescoping series 2 k k1 713515 k at 39 ml m S 39 I AI H I 31 51 I L 1 l39 339 S S 4 XEKVII J i T5 I 5 i w IM Sn I D VP I i K 3 3 a Link m m Page 4 of 8 Introduction to telescoping and geometric series Introductory example of a Geometric Series Page 5 of 8 B J Find thevalueof kl bt b 6lt 39 7 3 2 g 1 2 4 3 J i 39I quot 4 quotquot 4 439 nI 3 539 5 Squot g S i J 3a 19 232 3 Sn El 4 3 1 I 39 39 4 pFL S S J irn 6 0 6 4 quot39 4 0 I O in 5 y 39 quot t S S l I r 3 I r 7T 5 mvuducuun m teteswphg aha Eeumemc SEHES De nition and Theurem r Geometric Series A series whose ratio of successive mms s constant is called a gcomltric am series Further mur2 tf a The semes dwerges tf r21 The sertes cunverges tf t lt1 If the sertes cunverges then the sum at the sertes ts 1L where as the rst term uf the sertes er Examptes shawthattheserestsgeametrteahdthathesumttttextsts i Z 7M 4 n s 4 we ampw s l J39J 1quot n k you 3 11 F 39e n39 3 4 I k 55 H 393 J J quot39M h 2quot 39 d L kum wig4 Heinz1 a S 1qu 39 S4 l0 7 Vu i quot Ht Pagesm no 22k11 Find The sum if iT exisTs 2 k1 43 Show ThaT The series 2 4 1 is noT geomeTric 171 Introduction to telescoping and geometric series k quot 1 Show ThaT The series 2 1 is noT geomeTric k1 Page 7 of 8 Introduction to telescoping and geometric series Use an appropriate Geometric Series to determine a fractional form for the repeating decimal 2317317317 Page 8 of 8


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