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ID. nui~ber~ Math 2250-1 FINAL EXAM December 10, 2012 This exam is closed-book and closed-note. You may use a scientific calculator, but not one which is capable of graphing or of solving differential or linear algebra equations. Laplace Transform tables are included with this exam. In order to receive full or partial credit on any problem, you must show all of your work and jnsti1~’ your coilciusions. This exam counts for 30% of your course grade. It has been written so that there are 150 points possible, and the point values for each problem are indicated in the right-hand margin. Good Luck! problem score 1 2 possible 20 ______ 10 3 20 4 15 5 10 6 15 7 20 8 15 9 25 total 150 at rest at time t =jj4ec, is accelerated in a straight path provides a consta~force, and which is also subject to drag forces velocity. The boat Ic). Use your mo value problem engine that 2ONfor each mas 400 k ling abili to show that the boat velocity v(t) (in per second) satisfies the initial v’(t)2—.05v v(0)0 t\J&141k 2~ ~- f~as ,<.“ +uov / ~C•CD -ZOv ‘2- 1k). (5 points) c — 2 —. oSv Solve the initial value problem in i& V j- .o5y &oSt(vt÷ .o3v) e .o5t C (10 points) 2 •j 2-o5t44 v 2 e 4-~ 4cc v-t ~ Ct.t0 Ic). How far does the boat travel in the first 20 seconds? 12.0 ~((t0’) — C -Jo 0 9ot i- qv—4te --oct 4 (5 points) -~o5t-1zo ~ooe -Jo ttc$D€ 2 ,ç Here is a matrix and its reduced row echelon form: 3 9 -5 1 0 —6 1 2 -2 3 -2 0 tzt;f.r 13020 0 reduced row echelon form of A: 0 Find the general solution to the homogeneous matrix equation A&= combination form. ~.. 001 1o 00000 Write your solution in linear (6points) )f2t. ~tit r r -4: 4: k—~ 4EN I~ I ‘~3 + r 1-3 ‘I ~0 t1pa. V Let b be the first colunm ofA, i.e. 3 -2 1 I What is the general solution to the non-homogeneous equation A x= b? Hint: since you afready know the homogeneous solution from part g, you only need to find a particular solution and superposition in order to deduce your answer. Since a matrix times a vector is just a linear combination of the columns, and since you want A 410 be the first column of A, there is a natural choice for 4. r3 , -s 2 ~ I -z ~es%tJL o (3, L’J -J S.o X 1’ Jo (4 points) P Ia 0 r€k~ 3 3). Consider the following initial value problem, which could arise from Newtons second law in a forced mass-spring oscillation problem: x ~ ~f ‘h x~t1 x x(0)=0 x’ (0) = 6. — — 3~). If the applied force in this problem is 100 N, then what are the corr sponding values for the Hookes constant, mass, and damping coefficients? Include correct units. ~ CX’4- k~t ~ t~”~ F~ Ic-p 4-ti-. (5points) ~.xf4~g=5~ ≤O 4 c~ ~i2~ ~ 32 K’. 3 ktso 3~ Solve this initial problem using the methods of Chapter 5, based on particular and homogeneous solutions. 2 C) (15 points) ‘1 rti—G~- i-25 t(r+s) 4-i~ nt.4t (rf3)t__IG -~) r =) -h~ ,ç~(4’~t f1€~c.S4& tt4t~ r~-it4c 4~cii3t≤is.4faiJ =Ot-OI-25C. ;c7tc tSO ~)Ct2. Xj)t X)j I 1e~~os4f X(o)tOt2\ti~ c~=—2. ~,= o~3c1t4c~ ~) ç~-o 4 41 Re-solve the initial value problem in a using Laplace transfonns: x’’(t) +6x’(t) +25x(t)=50 x(O)=O x’ (0) = 6. (15 points) ?~X(s’i sKL5)-o) —øs--~ -I- 1~ S So+~s S Soi-GS ≤ +2SX(C~= ~P. Esi-C (s1.4~sns~ F’(st+c,stis’) i-($~pi-c)s 5o zs~~ SD+Gst 2(s~4-~s42S)#(&S~-)S t StUi~B) 4+ 1. (cc) ~) Ot24~3 d-~— 2 — 4- ~ - ~? B,—’~ ..__— 2i3tc-os4E 5 Use Lapiace transforms to solve the initial value problem x” (t) + w~x(t) = ~cos(w0 t) x(O) =x0 x’ (0) = 2) 0 ‘F c.~X(s) S ff_ t n (7points) ftc 4- ‘i” N(s~t f?~ S S St0 - ‘4 -f4- V0 ‘2. u. What physical phenomenon is exhibited by solutions to (3 points) kas a- 4eAIl~. t.aft. “4-i’wj. I, “~“fl ~“&~k -i-LI- 4~ e-~’~s / U DCLI.trs \ Cr. •swLLo.%eb. 4t~ cy.ki..~ik ~f ,4-c 42o~ct~c 6 ~i Find the general solution to the first order system of differential equations x1’(t) _3 I xI x21(t) 2 —2 x2 (10 points) Hint: The eigenvalues of the matrix are negative integers. rCA) (~+~)Cafz~-a ‘I ‘),t-cf,~ -2~ 2- ~io I —I 1.0 t>~+S~~+c~ to F ~‘1(t~ L’iii ate [!j b1rLt tnj (A4~1)~3’ —t -it Ltj L1e L~] ~ Sketch the phase portrait for this linear hoi ogeneous first order system, based on your work and eigendata in part a. Classify the equilbrium point at the origin. S~hu (0 X~ ,‘X~ 3 s44(t ki.4.j 2 3 2 1~~tt~ —11fl i~c~ I Lti 1-’ ~ce I 61~ )O 4asJac.s 7 I-, oo, So 5d4~4 appw~cL v~jii l3~~ ita fmsL’a”s t-) —02 ≤. .a~ fl Consider a general input-output model with two compartments as indicated below. The compartments contain volumes V~, V~ and solute amounts x1 (t), x2 (t) respectively. The flow rates (volume per time) are indicated by p., 1 1 .6. The two input concentrations (solute amount per volume) are c1, c5. . ~ 140 X~t~) ~ Supposer2 = p3 = = 100, r~ = r4 200, r5 = ‘~a~4I 0 . Explain whythevolumes V~(t), V2(t) remain constant. \(1’(4’j~zou +Ifl-jfl--~j~- a V1t~c.s~~4— V~canf. ≤o kD 2~~Using the flow rates above, c1 = 0.6, c5 = _%~y, 0 V~ = = (4points) 100 gal, show that the amounts of solutex1(t) intankl andx2(r) intank2satis~’ x1’(t) x2’(t) -3 = 1 2 —2 X1 x2 + 120 0 (6 points) / — r0 L~, ‘~n•(.~.)÷Iø~cs —3it~j V7. V1 ~‘LOj- ~z r~Lc—r0co Zn ‘5± .-2.OO ‘½ .. V1 So, ~- VT iz.c V 8 2k). Solve the initial value problem for the system ~10~t~° [x~’(t) = -3 1 X1 + 2 -2 assuming there is initi 1 no solute in either vector, and then use x = + o so e hj 120 0 Hint: Find a particular solution which is a constant P. Notice that you have already found iii problem 6. (10 points) —J Xp- çc~ 3 ~0 Lç 101.: [-s La] 1z o o CI ~Cz E~- E,-~7 2ztl.o fliti 4 -tj[c [no Lo 2c~ i-Ct 0 a)c1 .a~ ~çt S e, -~ _, t VT, #~xIi _, —91 tçso 4-st LGo — 1_I-I boi-c.14z ~ 4Zc1~Lt [Go L&o i Qt ~‘- aS~tojb~c~~ çx1w~ qo .4 L’o c.c. C I ~-qt 2 -t -4ff ~O ~Ljoe -2-at ~c,.L+) ~6o _to~;.t t20t~~ ~j4’~t. Qt ~ 9 Find the general solution to this second order system of differential equations, which could arise when modeling a coupled mass-spring system. Hint: You have already computed eigendata for the relevant matrix in problem x1’’(t) =—3x1 +x2 ~.. ‘‘(1) = 2 At 1-~ — 2 (c~ctfçs.k~) -t ‘~ C1t..s(+-.ej 1- .1 ~.kl • ~———~ II (6 points) ~ L’J + (L3L~ctb+%5(~Lt)f L ‘ Describe the two fundamental modes of vibration for this spring system. 5jew- 1%~444 k~asçec (C3t LqtD~ ~ac~ • fotsf Cctc~toN ~ L’iwA.L ~- si~n~p(s. k~asy, r~”~ ~ Lan~esa~ ~ i~, (4 points) fwcc.cq4f ~ s~$ ‘j, pkas.c, tj~4 n~pLi~.4~, 6ot2 ~ For Hooke’s constants and masses as shown below, show that the displacement functions x1 (t), x2 (t) of the two masses below (from their equilibrium hanging positions) satisfy the system of differential equations in this problem. I~_, I — 2x~”: I ..i rr t — Mx)#2Cs~x%)_ 2. (5 points) ‘~1~2r1 t~v: 2 —ts (.Kz~Xi’I N’— I, ~ 10 Ii We have studied and returned to the rigid rod pendulum several times in this course. This is the freelyrotating configuration indicated in the diagram below: 6’ L Lc~ 6 Using conservation of energy we have derived the autonomous second order differential equation that describes the angle e ( t) of the mass from the vertical reference line, at time 4 arriving at 0’’(r) + fsin(oQ)) =0. (i~ That second order DE above is equivalent to the autonomous first order system of two differential equations x’(t) 1 U) y’ (t) =- fsirqx) Explain this equivalence. ~W) £i€!rf(()~ ~u4~’~.a tkt~1.. (4points) -= e’ (4) i.4 {sIjn~c 4L~4 (a) C 2tOk1V~$tI~) 4 x(t), £oive (a’), TL, e’~~1 €≠‘~ So s~ j~i~i.*. 9”•+ sCk&to (z~ Find all equilibrium (i.e. constant) solutions to the first order system of bE’s above. Explain how these equilibrium solutions are related to the constant solutions of the second order differential equation for rigid-rod pendulum. 2U (x,,y~’l tthAt~•) ~- (6 points) ~ 3~’~z _tS(k1~x -~ ~‘≤CktxtO cc is C~ ~ ki- • fl~’~,Lt ~rL sA~- 44 k4Sh 4+ E’o+4-m— LA)”4b~- h.~s - 044 is. h.mcj .r~ ~sbh4$5 (0, hit) I kt&~ “S % lit -fn-- ~4K’th7t (.R,p) ~fr~4fl~ (Sal t3{ a4 t “C (2ti,o) 11 e e:~rr, h system repeated for your convenience: x’(t) y y’(t) =-fsin(x) ~ Use linearization and the Jacobian matrix to classit~’ the equilbriurn solutions to the first order system above. Indicate what you can deduce about the stability of these equilibrium solutions based only on the linearization. E(g, . [o -.~_ I Ht”~” (lOpoints) a]. tthei~t~..S)r(h%7t,o)t r~ Lt ~ oJ 2 -t . “4 noab, ~+~ta~I=t4 hoet4 KtktTt, ~mn tID ~tsj ~v~ai~ s1-o4~9.. I Lto lEXT~t~~ ) ~ Ii €~ ~(et t44& (n.4 Ci~Aa ‘lr~~;”t~ (kn,&~ ~ 4~ )~±y~ _____ 9~). Use the sep~ahön of v iables method we discussed in class (or if you prefer, use conservation of energy), to show that the parametric solution curves (x(t), y(1)) to the first order system above lie on the level curves for a certain function. Explain why this shows that the equilibrium solutions that are borderline based on the linearization work in ~, are actually stable for the original non-linear system. (5 points) 1k s~fk ~t Lx ‘:2s 0~12~~ .1. Lcd ~a So 7 ~ L’o-uts cos~c .1- C- _tcesxtc,(n4 vsot4n.. Cu.n’t5 a..a (ik~ f&ciç ~?, Lt&nts~ a≠,~ç~ ELx,& -~ — i~c a “~S+*n+ ‘kh.fl 44t I<C t~PE f,t ‘144 t Tkas EL~e.a) kaç5LoloJwAi~t. — ‘n4j.,* ~r ~ ~ ct~r yi5 ~ (km, h 12 €bfini ~e £fre’IftS~W ,41E. ~ So k%%L%S+ a.- - Table of Laplace Transforms This table summarizes the general properties of Laplace transforms and the Laplace transforms of particular functions derived in Chapter 10. fQ) F(s) af(t) ± bgQ) aF(s) + hG(s) f’Q) sF(s)—f(O) f”U) s2F(s) — sf(O) ?F(s) — s”~f(0) f (s — [(0) — .. . — sink: f~”(O) coshkt s sinhk: F(s — a) e~’ cask: (s —a)2±k2 (s a)JQ — a) r°’F(s) e°’ sink? ~~r~ga — r) di F(s)G(s) ~$sinkt — )n+l coskt f(r)dr e~’f(:) uQ adl :fQ) —F’(s) :~fQ) (—iyF’~~(s) f fit), periodp — ki coskt) sink: u(t—a) “ a a)2 ±k2 (2 k2)2 (82 _1_ k2)2 ~(sinkr±krcoskt) F(a)diy 1 — l—e~ — — - 8Q —a) e~’fQ)d1 - 1:/i (_j)L ,a, C” (square wave) 5 lFt1l — II (staircase) lLaJj I tanh as — s 2 — C” s(l — (‘I I I r(a ± I) 50+1 13