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Sample.TermTest1S08.Sol

Sample.TermTest1S08.Sol - UNIVERSITY OF WATERLoo TERM...

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Unformatted text preview: UNIVERSITY OF WATERLoo TERM TEST #1 SPRING TERM 2008 COURSE AM 250 COURSE TITLE INTRODUCTION TO DIFFEREN— TIAL EQUATIONS SECTION(S) 001 HELD WITH COURSE(S) N/A SECTION(S) OF HELD WITH COURSE(S) N/A DATE OF EXAM June 9th, 2008 TIME PERIOD 4:30 - 5:20 pm DURATION or EXAM NUMBER OF EXAM PAGES (including this cover sheet) INSTRUCTOR D. Harmsworth EXAM TYPE Closed book ADDITIONAL MATERIALS ALLOWED Scientific Calculator Name (print) ID Number Signature Instructions: 1. Print your name and ID Number on this page and sign it. 2. Answer the questions in the space pro— MARKING SCHEME Vided. Continue on the back of the page if necessary. Show all your work. 3. The last page may be removed and used for rough work. 4. Your grade will be influenced by how clearly you express your ideas and by how well you organize your solutions. AM 250 - Term Test #1 Spring Term 2008 Lfl [W] 1. (a) Find the general solution to the DE fig + 3mg = 6. [951 w dx db}: is \\mm(. 1m steals-r) how we \hme. (L9 3' 2133951 Q) A?!» l x, . 7> , :W\ an %\23m¥»3$n&u 0; :bdi leél: 639”)“ z ’35, 9* jaweofnxtkfl \\)J W lflowfi, 13:3 4, ‘53:) : 69L 3 1 i (b) Produce a qualitative sketch of the family of solutions, using information from both the DE and your result from part (a). Indicate as a dotted curve the set of points at which the slope of the solution curves is zero, and include some typical solution curves for which the constant of integration is positive, and some for which it is negative. Hint: as you should be able to see from your result to part (a), the solutions are all odd, so you can consider just a: > O to begin, and then use symmetry. (cub (fiber golmkm «fermek M‘s w. 1—7 ()0 ) @ slope, £130.90 Later-W éqo) EchfifionA Solvdguw. 3: ~31: («VJW L} 21"0 .‘zloec '- 31,3: 6) Le, \32 if :2. A“ §d\V~\“\’“3 "’7’ 00 5 a: O <9 g7}! Leasiaf’m} lnga‘» W» ”ix—3 o‘ 1‘ C70 b‘ * M 1') w or w) -— 00 (Q C4 0 I- : C >0 (9 5;, (auxin?) HA") 91m. SW 3“. do); ‘w 22W: * _ a; )(L-"TJ C) ( i$ 77 l, ' l ,. ‘ 3L 9‘“ (g) £th Lombi‘nl‘wf) all WE) twbcmhw im =0 M skew“ AM 250 - Term Test #1 Page 3 of 6 Spring Term 2008 2. Consider a circuit consisting of a resistor of resistance R, an inductor of inductance L, and a source voltage V(t). An inductor is a device which takes advantage of the fact that a current passing through a wire induces a magnetic field7 which can be magnified by wrapping the wire in a coil. However, for our purposes all you need to know is that the voltage drop across . . . d' . . such a dev1ce is given by VL(t) 2 LE; where 1(t) is the current (a formula confirmed by experiment). Also, the voltage across the resistor is VR(t) = Ri(t),’and these voltage drops must be balanced by the source voltage: di L— ' = V t . dt + R2 ( ) [2] a) Find the dimensions of R and L, given that the dimensions of voltage are M LQT‘ZQ‘1 (voltage is defined as the work required to move a charge from a reference point (where V = 0) to the given location, so it has dimensions of M — MM) [charge] — [charge] [filzCVMl : ("Wail—[Oq -.—. MLQT'\Q-; (Q Ct} 0“" CL}: {WU : MCT’O" : M905} Q) (Lat/3t} Q‘T} [3] b) Suppose V(t) = Voe‘kt. Define dimensionless variables as 7' = ELI and 17 = 9—3., and nondimensionalize the DE. ~Hc Lit, _‘, i2L : \[J D at Luau: v (Esta (:9 (it MAI At (Ld‘CL LAT -kLI 3'7 \[Cg [Vfl'IVoe ’K M “Eat ’31 if. if: 4’ ”z. t: 8 Q) [9] c) Solve the nondimensionalized DE, assuming that 1(0) = O and that 19 7é %. Then express yOur solution in terms of the original variables i and t. :5 . 7%: .CQ ALL-t ”E g, 41?) [,3 "1?: Aefi n Q “It 4”“ 3) ”if: 'A—‘Y‘ELQ P“ KL) ,ch \ V -u t It _. I C it m Salem“ Alf} 7* wL/lei € jg) ‘1 1 m l a) - l [L \ kL C o A ’ M 2: "” AO'E] [ I 5 [‘EL \Z-ch {l R. ’liL‘E :7 : 't .K; i 4‘ , m ’ 4’ I}, [)1 CC 4’ [L-[cl 8 fig) l‘” L” (Mi/{3 F - , t L1 c LE i Z? L(0]'0 llw‘ 42(0),]:9) $0 0 Cl 17H“ ) so Wk]. 0 . -ld— : E. a 7i , 8" [24¢ t TLEA T) ‘5 J} ['%<%] ~35: V0 - [L-(L (‘3 ’ Q AM 250 — Term Test #1 Page 4 of 6 Spring Term 2008 [6] e) What happens if k = %? Find the solution for this case. (Use the nondimensionalized DE, and then express your result in terms of the original variables.) 1" k9% ) SAN?» “v, AonfiWShOacl‘itéx BE l) ”‘C a fl *va ‘ 8 ‘ CD as ‘ New. @3an W7“; (’81 So WE weak 7P: ‘A’Eét Q?) ”(lake =0 Go 0) -‘C 5° ’7: [8 Ju— r“ LL L: Vi E:)€'% I l-Q- i: Elite L . (f) A _ r vkt Jit fl kkfih L” E<€ ”fl . 4; $1M» H30) 5° “'2 yf‘; El-eLtl llwew, surf-$9.3 L~7o, (g'lgfl-£, 1” we ) * A.“ M 'Et - \L, ' -E M 3‘: k77% “v" (.75 REE; “€1le IleLtve’ E ) {f Ml * Com slam.) WA “we Mock Otcwb d f AM 250 — Term Test #1 Spring Term 2008 [10] 3. For a celestial body in orbit around another (eg a moon orbiting a planet, or a satellite orbiting the earth) the period P of the orbit depends on the masses of the bodies, M1 and M2, the universal gravitational constant G (whose dimensions are [G] 2 M“1L3T‘2), and the semi—major axis of the ellipse of orbit, a (i.e. the distance between the objects, if the orbit is circular). Use Buckingham’s H Theorem to investigate how the period of a satellite depends on its distance from the earth. If one satellite is placed twice as high up as another, of the same mass, how much longer will it take to orbit the earth? VGWa VV<\D\A\£S‘A P1 { f3] 1 T \1“ Mi, [mm X} Ml, [Ma-)2 M ; < (D G , {a}: NET a {a}: L. J I «We NM N‘g VafiABtS) 61") F: 3 ((Jlafiuualw Sham 6apmm\elj). ’ lac) , kg) :73 MWQ; Cow Comfituu} l Mack/$3-; AWMWA‘CSI Van Tsfl ($7 0M E (ifllré‘ws " (“6‘ Vet k 030w, Mic M [mafili L314: 3 a i i ) MA So (g LJQ mdkflj “W, ‘33 ? (in) Alflk ‘73 0 UK, ‘E W; 7 V-B'G‘g'fi’j 75 [A30 AWMA‘ffis. @ ...
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