Sec 2.3 Modeling with First Order Equations

Transcription

1 Sc.3 Modling with First Ordr Equations Mathmatical modls charactriz physical systms, oftn using diffrntial quations. Modl Construction: Translating physical situation into mathmatical trms. Clarly stat physical principls blivd to govrn procss. Diffrntial quation is a mathmatical modl of procss, typically an approximation. Analysis of Modl: Solving quations or obtaining qualitativ undrstanding of solution. May simplify modl, as long as physical ssntials ar prsrvd. Comparison with Exprimnt or Obsrvation: Vrifis solution or suggsts rfinmnt of modl.

2 Exampl 1: Salt Solution (1 of 11) (Ex 1) At tim t =, a tank contains 5 lb of salt dissolvd in 1 gal of (sal watr. Assum that watr containing ¼ lb of salt/gal is ntring tank at rat of r gal/min, and lavs at sam rat. (a) St up IVP that dscribs this salt solution flow procss. (b) Find amount of salt Q( in tank at any givn tim t. (c) Find limiting amount Q L of salt Q( in tank aftr a vry long tim.

3 Exampl 1: Salt Solution ( of 11) (Ex 1) At tim t =, a tank contains 5 lb of salt dissolvd in 1 gal of (sal watr. Assum that watr containing ¼ lb of salt/gal is ntring tank at rat of r gal/min, and lavs at sam rat. Q( = th amount (lb) of th salt in th tank at tim t. dq/dt = th rat of chang of Q( (a) St up IVP that dscribs this salt solution flow procss. (b) Find amount of salt Q( in tank at any givn tim t.

4 Exampl 1: Salt Solution (3 of 11) (Ex 1) At tim t =, a tank contains 5 lb of salt dissolvd in 1 gal of (sal watr. Assum that watr containing ¼ lb of salt/gal is ntring tank at rat of r gal/min, and lavs at sam rat. (c) Find limiting amount Q L of salt Q( in tank aftr a vry long tim.

5 Exampl : Salt Solution (4 of 11) (Ex ) At tim t =, a tank contains Q lb of salt dissolvd in 1 gal of (sal watr. Assum that watr containing ¼ lb of salt/gal is ntring tank at rat of r gal/min, and lavs at sam rat. (a) St up IVP that dscribs this salt solution flow procss. (b) Find amount of salt Q( in tank at any givn tim t. (c) Find limiting amount Q L of salt Q( in tank aftr a vry long tim. (d) If r = 3 & Q = Q L, find tim T aftr which salt is within % of Q L. () In (d) find flow rat r rquird if T is not to xcd 45 min.

6 Exampl : (a) Initial Valu Problm (5 of 11) At tim t =, a tank contains Q lb of salt dissolvd in 1 gal of watr. Assum watr containing ¼ lb of salt/gal ntrs tank at rat of r gal/min, and lavs at sam rat. Assum salt is nithr cratd or dstroyd in tank, and distribution of salt in tank is uniform (stirrd). Thn dq / dt ratin - ratout Rat in: (1/4 lb salt/gal)(r gal/min) = (r/4) lb/min Rat out: If thr is Q( lbs salt in tank at tim t, thn concntration of salt is Q( lb/1 gal, and it flows out at rat of [Q(r/1] lb/min. Thus our IVP is dq dt r 4 rq, 1 Q() Q

7 Exampl : (b) Find Solution Q( (6 of 11) To find amount of salt Q( in tank at any givn tim t, w nd to solv th initial valu problm dq dt rq 1 Q() To solv, w us th mthod of intgrating factors: r, 4 Q ( Q( Q( at rt/1 5 rt/1 rt/ 1 Q 5 r 4 rt/1 dt rt/1 5 rt/1 C 5 C rt/1 or Q( 5 rt/1 rt/ 1 1 Q

8 Exampl : Salt Solution (7 of 11) (Ex ) At tim t =, a tank contains Q lb of salt dissolvd in 1 gal of (sal watr. Assum that watr containing ¼ lb of salt/gal is ntring tank at rat of r gal/min, and lavs at sam rat. (a) St up IVP that dscribs this salt solution flow procss. (b) Find amount of salt Q( in tank at any givn tim t. (c) Find limiting amount Q L of salt Q( in tank aftr a vry long tim. (d) If r = 3 & Q = Q L, find tim T aftr which salt is within % of Q L. () In (d) find flow rat r rquird if T is not to xcd 45 min

9 Exampl : (c) Find Limiting Amount Q L (8 of 11) Nxt, w find th limiting amount Q L of salt Q( in tank aftr a vry long tim: rt/1 Q lim Q( lim 5 Q 5 5 L t t This rsult maks sns, sinc ovr tim th incoming salt solution will rplac original salt solution in tank. Sinc incoming solution contains.5 lb salt / gal, and tank is 1 gal, vntually tank will contain 5 lb salt. lb Th graph shows intgral curvs for r = 3 and diffrnt valus of Q. Q( 5 rt/1 rt/ 1 1 Q

10 Exampl : (d) Find Tim T (9 of 11) Suppos r = 3 and Q = Q L. To find tim T aftr which Q( is within % of Q L, first not Q = Q L = 5 lb, hnc Nxt, % of 5 lb is.5 lb, and thus w solv Q( 5 rt/1.3t Q T..3T ln(.).3t ln(.) T 13.4 min.3

11 Exampl : () Find Flow Rat (1 of 11) To find flow rat r rquird if T is not to xcd 45 minuts, rcall from part (d) that Q = Q L = 5 lb, with rt/1 Q( and solution curvs dcras from 5 to 5.5. Thus w solv r. 45 r 1 ln(.).45r ln(.) r 8.69 gal/min.45

12 Exampl : Discussion (11 of 11) As long as flow rats ar accurat, and concntration of salt in tank is uniform, thn diffrntial quation is accurat dscription of flow procss. Modls of this kind ar oftn usd for pollution in lak, drug concntration in organ, tc. Flow rats may b hardr to dtrmin, or may b variabl, and concntration may not b uniform. Also, rats of inflow and outflow may not b sam, so variation in amount of liquid must b takn into account. Sinc situation is hypothtical, th modl is valid.

13 Exampl : Compound Intrst In gnral, if intrst in an account is to b compoundd m tims a yar, rathr than continuously, th quation dscribing th amount in th account for any tim t, masurd in yars, bcoms: S ( S(1 r / m) Th rlationship btwn ths two rsults is clarifid if w rcall from calculus that mt lim mt S (1 r / m) S m rt Growth of Capital at a Rturn Rat of r = 8% For Svral Mods of Compounding: S(/S() t m = 4 m = 365 xp(r Yars Compound d Quartrly Compound d Daily Compound d Continuous ly A comparison of th accumulation of funds for quartrly, daily, and continuous compounding is shown for short-trm and long-trm priods.

14 (Ex ) If a sum of mony is dpositd in a bank that pays intrst at an annual rat, r, compoundd continuously, (1) Find th ODE that th amount of mony S = S( at any tim in th fund will satisfy: () Solv th ODE with th initial amount S() = S

15 Exampl : Compound Intrst ( of 3) If a sum of mony is dpositd in a bank that pays intrst at an annual rat, r, compoundd continuously, th amount of mony S = S( at any tim in th fund will satisfy th diffrntial quation: ds dt rs, S() S whr S rprsnts th initial invstmnt. Th solution to this diffrntial quation, found by sparating th variabls and solving for S, bcoms: rt S( S, whr t is masurd in yars Thus, with continuous compounding, th amount in th account grows xponntially ovr tim.

16 Exampl : Dposits and Withdrawals (3 of 3) Rturning now to th cas of continuous compounding, lt us suppos that thr may b dposits or withdrawals in addition to th accrual of intrst, dividnds, or capital gains. If w assum that th dposits or withdrawals tak plac at a constant rat k, this is dscribd by th diffrntial quation: ds dt rs k or in standard form rs and S() whr k is positiv for dposits and ngativ for withdrawals. W can solv this as a gnral linar quation to arriv at th solution: To apply this quation, suppos that on opns an IRA at ag 5 and maks annual invstmnts of $ thraftr with r = 8 %. ds dt rt rt S( S ( k / r)( 1) k S At ag 65, S(4) *.8*4 (/.8)(.8*4 1) $588,313

17 Exampl 3: Pond Pollution (1 of 7) Considr a pond that initially contains 1 million gallons of frsh watr. Watr containing toxic wast flows into th pond at th rat of 5 million gal/yar, and xits at sam rat. Th concntration c( of toxic wast in th incoming watr varis priodically with tim: c( = + sin t (g/gal) (a) Construct a mathmatical modl of this flow procss and dtrmin th amount Q( of toxic wast in pond at tim t. (b) Plot solution and dscrib in words th ffct of th variation in th incoming concntration.

18 Exampl 3: (a) Initial Valu Problm ( of 7) Pond initially contains 1 million gallons of frsh watr. Watr containing toxic wast flows into pond at rat of 5 million gal/yar, and xits pond at sam rat. Concntration is c( = + sin t g/gal of toxic wast in incoming watr. Assum toxic wast is nithr cratd or dstroyd in pond, and distribution of toxic wast in pond is uniform (stirrd). Thn dq / dt ratin - ratout Rat in: ( + sin t g/gal)(5 x 1 6 gal/yar) Rat out: If thr is Q( g of toxic wast in pond at tim t, thn concntration of salt is Q( lb/1 7 gal, and it flows out at rat of [Q( g/1 7 gal][5 x 1 6 gal/yar]

19 Exampl 3: (a) Initial Valu Problm, Scaling (3 of 7) Rcall from prvious slid that Rat in: ( + sin t g/gal)(5 x 1 6 gal/yar) Rat out: [Q( g/1 7 gal][5 x 1 6 gal/yar] = Q(/ g/yr. Thn initial valu problm is dq dt Q( 6 sin t 5 x 1, Q() Chang of variabl (scaling): Lt q( = Q(/1 6. Thn dq dt q( 1 5sin t, q()

20 Exampl 3: (a) Solv Initial Valu Problm (4 of 7) To solv th initial valu problm q q / 1 5sin t, q() w us th mthod of intgrating factors: ( q( at t / t / t / 1 5sin t Using intgration by parts (s nxt slid for dtails) and th initial condition, w obtain aftr simplifying, q( q( t / 4 17 t / cos t 4 17 t / dt cos t 1 sin t t / t / sin t C

21 Exampl 3: (a) Intgration by Parts (5 of 7) t / 5 sin tdt t / t / t / sin tdt sin tdt sin tdt t / t / t / 1 t / cos t cos t cos t t / t / t / t / cos tdt t / sin t 1 t / cos t sin t C 8 t / cos t sin t C 17 1 t / cos t sin t C 17 sin t t / t / sin tdt sin tdt

22 Exampl 3: (b) Analysis of solution (6 of 7) Thus our initial valu problm and solution is dq dt q( 1 q 1 5sin t, 4 17 cos t 1 17 q() sin t 3 17 t / A graph of solution along with dirction fild for diffrntial quation is givn blow. Not that xponntial trm is important for small t, but dcays away for larg t. Also, y = would b quilibrium solution if not for sin( trm.

23 Exampl 3: (b) Analysis of Assumptions (7 of 7) Amount of watr in pond controlld ntirly by rats of flow, and non is lost by vaporation or spag into ground, or gaind by rainfall, tc. Amount of pollution in pond controlld ntirly by rats of flow, and non is lost by vaporation, spag into ground, dilutd by rainfall, absorbd by fish, plants or othr organisms, tc. Distribution of pollution throughout pond is uniform.

24 (Ex 4) A body of mass m is projctd away from th arth in a dirction prpndicular to th arth s surfac with initial vlocity v and no air rsistanc. Taking into account th variation of th arth s gravitational fild with distanc, th gravitational forc acting on th mass is mgr W( x) ( R x) : a function of x whr x is th distanc abov th arth s surfac. R is th radius of th arth and g is th acclration du to gravity at th arth s surfac. (1) St up an ODE for th vlocity, and solv it. () Find th maximum hight (h) of th body. (3) Find th limit of th initial vlocity as h.

25 Exampl 4: Escap Vlocity (1 of ) A body of mass m is projctd away from th arth in a dirction prpndicular to th arth s surfac with initial vlocity v and no air rsistanc. Taking into account th variation of th arth s gravitational fild with distanc, th gravitational forc acting on th mass is mgr w( x) whr x is th distanc abov th arth's surfac ( R x) R is th radius of th arth and g is th acclration du to gravity at th arth s surfac. Using Nwton s law F = ma, dv mgr m, v () v dt ( R x) dv dt dv dx dx dt dv v dx Sinc and canclling th m s, th diffrntial quation bcoms dv gr v, sinc x whn t, v( ) v dx ( R x)

26 Exampl 4: Escap Vlocity ( of ) dv gr v, v( ) v dx ( R x) W can solvr th diffrntial quation by sparating th variabls and intgrating to arriv at: gr Th maximum hight (altitud) will b rachd whn th vlocity is zro. Calling that maximum hight ξ, w hav W can now find th initial vlocity rquird to lift a body to a hight ξ : v gr R and, taking th limit as ξ, w gt th scap vlocity, rprsnting th initial vlocity rquird to scap arth s gravitational forc: Notic that this dos not dpnd on th mass of th body. v gr C R x gr R x v v gr v R gr v