Math Spring 2015 PRACTICE FINAL EXAM (modified from Math 2280 final exam, April 29, 2011)

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1 ame ID number Mah 8- Sring 5 PRACTICE FIAL EXAM (modified from Mah 8 final exam, Aril 9, ) This exam is closed-book and closed-noe You may use a scienific calculaor, bu no one which is caable of grahing or of solving differenial or linear algebra equaions Lalace Transform and inegral ables are included wih his exam In order o receive full or arial credi on any roblem, you mus show all of your work and jusify your conclusions This exam couns for 3% of your course grade I has been wrien so ha here are 5 oins ossible, and he oin values for each roblem are indicaed in he righ-hand margin Good Luck! roblem score ossible oal 5

2 ) Find he marix exonenials for he following wo marices Work one of roblems using he ower series definiion, and he oher one using he fundamenal marix soluion aroach (your choice) As i urns ou, boh mehods are reasonable for boh roblems a) A = ( oins) soluion via FM: e A = F F K : For F he columns will form a basis for he soluion sace o x#= A x E l = : E l =K : A K l I = Kl K K Kl = l K = l K l C v = v = K eigenvecor eigenvecor x H = c e C c ek K F = e e K e Ke K is an FM e A = F F K = e e K e Ke K K = e e K K K e Ke K = e e K K K e Ke K K = e C e K e Ke K e Ke K e C e K If you ry ower series you will ge enries which are he ower series for cosh, sinh (We did no review hose in his 8 class) In fac, an equivalen way o wrie e A in his case is e A cosh sinh = sinh cosh K b) wih ower series: Powers of B: B = B = K e B = I C B C! B CC n n! Bn C K, B = K K = K K = KI, B 3 = B B =KI B =KB, B 4 = B B = KI KI = I B 5 = B 4 B = B ( oins)

3 and he aern reeas cyclicly, every four owers Thus e B = C K C K! K C 3 3! K C 4 4! C = K! C 4 4! K C 3 3! K 5 5! C K 3 3! C 5 5! K C K! C 4 4! C = cos Ksin sin cos soluion via FM: e A = F F K : For F he columns will form a basis for he soluion sace o x#= A x E l = i : A K l I = Kl K Kl = l C = l K i l C i Ki K Ki v = eigenvecor i comlex soluion z = e i cos = cos C i sin = C i i i Ksin The real and comlex ars are each real soluions, so a FM is given by cos sin F = Ksin cos Since F = I, his is e A e A = cos Ksin sin cos sin cos a) Use Lalace ransform echniques o find he general soluion o he undamed forced oscillaor equaion wih resonance: x## C w x = F sin w soluion: The soluion x makes boh sides of he DE equal, so heir Lalace ransforms are oo x = F w = F w s X s K s x K v C w X s = F w X s s C w = F w X s = F w w 3 s C w s C w C s x C v s C w C x s C v C w s C w sin w K w cos w sin w K w cos w s C x cos w C C x cos w C v w sin w v w sin w ( oins) 3

4 b) Use Lalace ransform o find he general soluion o he non-resonan undamed forced oscillaor equaion x## C w x = F sin w w s w soluion: s w X s K s x K v C w X s = F s C w X s s w C w = F s C w C s x C v s X s = F w C x s C w s C w s C v C w s C w X s = F w w K w s K s C w s C w C x s C v C w s C w x = F w w K w w sin w K w sin w C x cos w C v w sin w ( oins) 3) Consider he following hree-ank configuraion Le ank i have volume V i and solue amoun x i a ime Well-mixed liquid flows beween anks one and wo, wih raes r, r, and also beween anks wo and hree, wih raes r 3, r 4, as indicaed 3a) Wha is he sysem of 6 firs order differenial equaions governing he volumes V, V, V 3 and solue amouns x,,? (Hin: Alhough mos of our recen anks have had consan volume, we've also discussed how o figure ou how fas volume is changing in inu/ouu models (6 oins) V # = r K r V # = r C r 4 K r K r 3 V 3 # = r 3 K r 4 x # =Kr x V C r V # = r x V K r C r 3 V C r 4 V 3 # = r 3 V K r 4 V 3 3b) Suose ha all four raes are gallons/hour, so ha he volumes in each ank remain consan Suose ha hese volumes are each gallons Show ha in his case, he differenial equaions in (a) 4

5 for he solue amouns reduce o he sysem x # # # = K K K soluion: In his case, each r i V i has numerical value = so he differenial equaions for he x j simlify o x # =Kr x V C r V =Kx C # = r x V K r C r 3 V C r 4 V 3 = x K C # = r 3 V K r 4 V 3 = K which can be rewrien in he marix vecor form dislayed above 3c) Male o he rescue! Male says ha x (4 oins) > > wih LinearAlgebra : A d Mari, 3, K,,,,K,,,,K ; Eigenvecors A ; A := K K K K K3, K K () Use his informaion o wrie he general soluion o he sysem in (3b) (5 oins) soluion: The eigenvalues are in he firs column of ouu, and he corresonding eigenvecors are in he columns of he marix Each eigenair l, v yields a soluion e l v so he general soluion o x# = A x is x = c C c e K K C c 3 e K3 K 3d) Solve he iniial value roblem for he ank roblem in (3b), assuming here are iniially ounds of solue in ank, ounds in ank, and none in ank 3 ( oins) Using he soluion above, a = : 5

6 KR C R /R, KR C R 3 /R 3 : = c C c K K K K3 K R C R 3 /R 3 K K3 6 K K K3 C c 3 K R 3 6 /R 3 K K3 K5 3 R 3 C R /R, KR 3 C R /R K 5 K5 K5 R C R /R so c =, c =K5, c 3 =K5 and K5 K5 x = CK5e K K CK5e K3 K 3e) Wha is he limiing amoun of sal in each ank, as aroaches infiniy? (Hin: You can deduce his answer, no maer wheher you acually solved 4d, bu his gives a way of arially checking your work here) (5 oins) Since here are 3 ounds oal of sal and since he anks each have he same volume, as he concenraions converge o heir final uniform concenraion, he sal amouns will converge o ounds er ank This is also clear from he soluion formula, as he second wo erms decay exonenially o zero 4) Alhough we usually use a mass-sring configuraion o give conex for sudying second order differenial equaions, he rigid-rod endulum also effecively exhibis several key ideas from his course 6

7 Recall ha in he undamed version of his configuraion, we le he endulum rod lengh be L, assume he rod is massless, and ha here is a mass m aached a he end on which he verical graviaional force acs wih force m$g This mass will swing in a circular arc of signed arclengh s = L$q from he verical, where q is he angle in radians from verical The configuraion is indicaed below 4a) Use he fac ha he undamed sysem is conservaive, o derive he differenial equaion for q, q## C g L $sin q = ( oins) Hin: Begin by exress he TE=KE+PE in erms of he funcion q and is derivaives Then comue TE# and se i equal o zero TE = KE C PE = m v C mgh Measure he arclengh s from he boom o he mass locaion, and i's given by s = L q The scalar velociy is v = s# = Lq# Measure heigh from he boom and i is given by h = L K L cos q Thus TE = m Lq# C m g L K L cos q Toal energy consan is equivalen o TE# h, ie h ml q# q## C mgl sin q q# h mlq# L q## C g sin q Since q# can only be zero a isolaed imes, i mus be ha h L q## C g sin q which is he same as he claimed DE, if we divide boh sides by L 4b) Exlain recisely how he second order differenial equaion in (5a) is relaed o he firs order sysem of differenial equaions y x' y' = K g sin x L (5 oins) 7

8 soluion: Le q solve q## C g L $sin q = Define x d q, y = q# Then x# = q# = y y# = q## =K g L sin q =Kg L sin x which is he dislayed sysem (Conversely, if x, y yields a soluion o he second order DE for q ) T solve he sysem, hen defining q d x 8

9 5) Consider he following 3-mass, -sring zero-drag "rain"configuraion below A res he cars are searaed by cerain disances and he srings are neiher ulling nor ushing From ha equilibrium configuraion, he rain is u ushed ino moion along a rack, and he dislacemens from equilbrium are measured by x,, as indicaed 5a) Use ewon's law and he Hooke's law (linearizaion), o derive he sysem of differenial equaions for x,, m x ## = k K x m ## =Kk K x C k K m 3 ## =Kk K (8 oins) 5b) Show ha in case unis are chosen so ha he numerical values of all he masses are he same as he numerical values of he sring, ie m = m = m 3 = k = k = k 3 hen he sysem above reduces o x ## ## ## = K K K x (4 oins) In his case we may divide each of he DE's in 5a by he corresonding mass, and relace each k j m j by This yields x ## = K x =Kx C ## =K K x C K = x K C ## =K K = K which is equivalen o he marix vecor sysem ha is dislayed 9

10 5c) Exhibi he general soluion for he sysem in 5b oe ha you've already seen his marix in roblem 3: > > wih LinearAlgebra : A d Mari, 3, K,,,,K,,,,K ; Eigenvecors A ; A := K K K K K3, K K () (8 oins) soluion: Recall (and you should be able o exlain why) ha if l, v is an eigenair for he marix A, wih l!, hen for w = Kl we ge soluions cos w v, sin w v If l = we ge soluions v, v So, x = c C c C c 3 cos C c 4 sin K C c 5 cos 3 C c 6 sin 3 K 5d) Describe he general moion of he rain as a suerosiion of hree fundamenal modes In he firs mode, c C c (5 oins) The rain is moving wihou oscillaion, having sared a c from he chosen origin, and wih velociy c In he second mode, c 3 cos C c 4 sin he firs and hird cars are oscillaing ou of hase and wih equal amliudes, while he second car remains saionary In he hird mode c 5 cos 3 C c 6 sin 3 he cars are oscillaing he mos raidly, wih he firs and hird cars in hase wih equal amliude, and he middle car ou of hase, wih wice he amliude of he ouer wo cars K K

11 6a) We consider a -eriodic saw-ooh funcion, given on he inerval (K, ) by f =, and equal o zero a every ineger mulile of Here's a grah of a iece of his funcion: Derive he Fourier series for f, 3 K8 K6 K4 K K K3 f = > n = K n C sin n n soluion: Because f is an odd funcion, is Fourier cosine coefficiens are all zero wih f w a C n > a n cos n = L C n > b n sin n = L = n > b n sin n = L b n d f, sin n L = K f sin n d = sin n d ( oins) (The las se holds because he inegrand is odd*odd=even funcion) Inegrae by ars, leing u =, du = d, dv = sin n, v =K n cos n sin n d = K n cos n = = K n cos n C n sin n K n cos n d =K n K n K = n K n C Seing b n = n K n C, a =, a n = yields he dislayed Fourier series f = > n = K n C sin n n 6b) Use he Fourier series above o exlain he ideniy 4 = K 3 C 5 K 7 C 9 K (5 oins) soluion Since f is differeniable a = he Fourier series converges o f = here, ie

12 Since sin n f = for n even, sin = = > n = n K n C sin n n =C for n = C 4 k, k ;, sin n = 3 C 4 k, k ;, and since K n C = for n odd, we ge = K 3 C 5 K Divide boh sides by o ge he dislayed ideniy n =K for 7) Consider he saw-ooh funcion f from roblem 6, and he forced oscillaion roblem x## C 9$x = f 7a) Discuss wheher or no resonance occurs (5 oins) The naural angular frequency is w = 3 Since he Fourier exansion of f has a sin 3 erm, here will be resonance 7b) Find a aricular soluion for his forced oscillaion roblem Hin: Use he Fourier series for f given in roblem 6 You may make use of he aricular soluions able on he nex age ( oins) We use (infinie) suerosiion o find a aricular soluion For n s 3 he forced oscillaion roblem has aricular soluion x## C 9$x = > n = x## C 9$x = n K n C sin n n K n C sin n = n K n C sin n 9 K n For n = 3 he forced oscillaion roblem has a aricular soluion Thus for we have a aricular soluion = 3 x## C 9$x = 3 sin 3 x## C 9$x = > n = K 6 cos 3 =K 9 cos 3 K n C sin n n =K n C 9 cos 3 C K n > sin n s 3 n 9 K n (The sum on he righ converges o a bounded funcion, wih absolue value less ha

13 by raio comarison o he convergen series > n s 3 n 9 K n! >n = n 3 ) 8) We discussed he analogy beween consan coefficien firs-order linear differenial equaions (in Chaer ), and firs order sysems of differenial equaions (In Chaer 5) Use marix exonenials and he "inegraing facor" echnique o show ha for firs order sysems wih consan marix A, he general soluion o x# = A x C f is given by he formula x = e A e K A f d C e A c (In he formula above, e K A f d is sanding for any aricular aniderivaive of e KA f, and he dislayed formula is exressing x as C x H ) Hin: begin by rewriing he sysem as x# KA x = f and hen find an aroriae (marix) inegraing facor ( oins) soluion: x# KA x = f e K A x# KA x = e K A f d d e K A x = e K A f e K A f = e K A f d C c x = e A e K A f d C e A c = C x H oe: One uses he "universal" roduc rule we discussed in class, and d d ek A = e K A A o jusify d e K A x =e K A x# KA x d 3

14 Fourier series informaion: For f of eriod P = L, wih f w a C n > a n cos n = L C n > b n sin n = L a = L a f d (so L = L f d is he average value of f) L KL KL a n d f, cos n L = L f cos n L L d, n ; KL b n d f, sin n L = L f sin n L L d, n ; KL Paricular soluions from Chaer 3 or Lalace ransform able: x## C w x = A sin w A = sin w when w s w w K w =K A cos w w when w = w x## C w x = A cos w = A cos w w K w when w s w = A sin w w when w = w wih wih x##c c x#c w x = A cos w c O = x s = C cos w K a A C = w K w C c w w K w cos a = w K w C c w c w sin a = w K w C c w x##c c x#c w x = A sin w c O = x s = C sin w K a A C = w K w C c w 4

15 cos a = sin a = w K w w K w C c w c w w K w C c w 5

16 6