CHAPTER 3. Section 1. Modeling Population Growth

Transcription

1 CHAPTER 3 Section 1. Modeling Population Gowth 1.1. The equation of the Malthusian model is Pt) = Ce t. Apply the initial condition P) = 1. Then 1 = Ce,oC = 1. Next apply the condition P1) = 3. Then 3 = 1e. Solving gives the epoductive ate = ln The population afte five days is P5) = 1e 5ln3 = The equation of the Malthusian model is Pt) = Ce t. Apply the cell counts to solve fo and C. P1) = 1, so Ce = 1, i.e. = ln1/c). Also, P2) = 3, so Ce 2 = 3, i.e. = 1 2 ln3/c). Setting these equal and solving, one obtains C = 1/3. Substituting C into = ln1/c) gives that = ln In addition P) = C = 1/ Recall the equation of the Malthusian model Pt) = Ce t. The constant C is the initial population P), so the tipling condition is stated as P1) = P)e 1 = 3P). Thus, = 1/1) ln The population doubles pecisely when Pt) = 2P) = P)e t ln 3)/1, that is, when t ln 3)/1 = ln 2. Solving, one obtains t = 1 ln 2)/ ln The solution of the Malthusian model is Pt) = Ce t. The constant C is the initial population P), so the doubling condition is stated as P1) = P)e 1 = 2P). Thus, = 1/1) ln 2. To find the value of t so that Pt) = 1, when P) = 1, solve 1 = 1e t ln 2)/1 fo t. This gives t = 1 ln 1)/ ln The modified Malthusian model will take the fom P = P h. Multiplying by the integating facto e t and putting the tems involving P on the left gives the equation P e t e t P = he t. Solve to obtain Pe t = h/)e t + C, whee C is a constant. This constant has the value P) h/), obtained by substituting t =. Multiply by e t and get the solution Pt) = h/) + P ) h/))e t. The long-time activity of the population depends entiely on the sign of P) h/). If it is negative, then the population will die out. If it is positive, the population will gow exponentially. If it is zeo, then the population will emain at a constant h/ Fist, calculate the epoduction ate. Unde nomal Malthusian conditions, the population doubles evey eight hous, so P8) = 2P) = P)e 8. Thus, = 1/8) ln Now, using the solution to the modified Malthusian model P = P h, obtain Pt) = h/) + P ) h/))e t. Hee, P) = 2 and h = 2, so Pt) = 16/ ln 2) / ln 2))e t ln 2)/ e.866t. The cultue will be depleted when 16/ ln 2) / ln 2))e t ln 2)/8 =. Solving fo t gives t = 8/ ln 2) ln 16/2 ln 2 16)) Now, if P) = 25, then the solution becomes Pt) = 16/ ln 2) / ln 2))e t ln 2)/ e.866t. When t is as above, Pt) Recalling the solution to poblem 3.1.5, one will note that this indeed should be the case, as the initial population P) is now geate than the citical value h/ The epoduction ate is calculated as befoe by solving P4) = 2P) = P)e 4. Thus, = 1/4) ln Using the esult fom poblem 3.1.5, the population is given by Pt) = h/) + P ) h/))e t. In ode fo the population to not explode, the coefficient P) h/) must not be positive. Hence the havesting ate must be at least P), which hee is 1)1/4) ln 2 = 25 ln

2 14 Chapte 3. Modeling and Applications 1.8. a) The fist data point is P) = 1, and the second is P1) = 14. The equation of the Malthusian model is Pt) = Ce t. Thus, C = P) = 1, and = lnp 1)/1) = ln7/5). See the following figue fo plots Malthusian data 2 flies time b) Now use the data points P) = 1 and P5) = 38. Using the equation of the Malthusian model, we get C = P) = 1 and = 1/5) lnp 5)/1) o = 1/5) ln3.8). See the following figue fo plots. 15 Malthusian data 1 flies time c) Using linea egession on the function ln P = ln C + t hee gives a system ) ) ) ln C = The solution to this is C and See the following figue fo plots. 15 Malthusian data 1 flies time d) The epoductive ate found using only the fist two data points pat a) was the least accuate. The ate found using an ealy data point and a late data point pat b) impoved the accuacy. The best estimate of came fom the linea egession in pat c, because it used all of the data.

3 3.1. Modeling Population Gowth 15 e) As long as an accuate value of is used, the Malthusian model is a good appoximation of the data in the table To find the minimum and maximum gowth ate, use the fist deivative test on the gowth ate dp /dt. That is, set d 2 P/dt 2 =. Obtain 1 2P k d 2 P dt = 2 P Thus, dp /dt eaches an extemum when P = and when P = k/2. One wants to show that dp /dt is a maximum when P = k/2. So, one checks that d 2 P/dt 2 changes sign fom positive to negative at P = k/2. When P = k/2, P = k/2)1 k/2)/k) = k/4, which is positive. So any possible change in sign of d 2 P/dt 2 is in the thid facto 1 2P/k). This facto does change sign fom positive to negative at P = k/2, so d 2 P/dt 2 changes sign fom positive to negative at P = k/2. Theefoe dp /dt eaches a maximum at P = k/ a) Let ω = αp and s = βt. Then solving fo P and t and substituting into the logistic equation, one obtains β dω α ds = ω 1 ω ). α αk Isolating dω/ds and simplifying, one obtains dω ds = β ω αβk ω2. This is as desied. b) If β = and α = 1/K, then we have dω/ds = ω ω 2. c) Sepaating vaiables and using patial factions, one obtains 1 ω 1 ) dω = ds. 1 ω Thus, ln ω ln 1 ω = s + c. Exponentiating and combining all constants into a constant a gives ω 1 ω = aes. Solving fo ω gives the desied fom d) Note that P = ωk and s = t. Thus, ω = 1 1 a)e s. ). K P = 1 + ae. t The initial condition Pt ) = P gives P + P ae t = K, so a = K P e t. P Using this in the equation fo P, one obtains KP P = P + K P )e. t t ) This is the solution 1.13) given in the text The caying capacity K = 2,, and the initial condition P = 1, and it is given that P1) = 2. Using equation 1.13), one obtains 2 = 2)1) )e 1.

4 16 Chapte 3. Modeling and Applications Solving gives = ln9/19))/ Afte 25 hous, the population is 2)1) P25) = )e2.5ln9/19) Now, find t so that Pt) = K/2 = 1. This gives e 1 1 ln 9 19 )t = That is, t = 1 ln19))/ln9/19)) The caying capacity K = 2,, and the initial condition P = 1, and it is given that P8) = 12. Using equation 1.13), one obtains 12 = 2)1) )e. 8 Solving gives = 1/8) ln188/12)19)).241. Now, find t so that Pt) = 3K/4 = 15. This gives e t = 1/57. That is, t = ln 57)/ a) Use the following thee data points P = P) = 6, P 1 = P9) = 52, and P 3 = P18) = 163 to estimate and k: Set h = 9. Equation 1.17) gives = 1 ) ) 9 ln = ) Then using this in equation 1.16), one obtains K = 6)52)1 e.2699) ) = e.2699) The plot of the logistic solution with these paametes is shown in the following figue. 4 3 P t b) To compute the paametes we want to find values of the paametes P,, and K which minimize 11 j=1 [ )] Ptj ) 2 log, P j whee t j and P j ae the data fom Table 1, and Pt) is the logistic function in equation with paametes P,, and K. Using a nonlinea least squaes minimization pogam we find that K =

5 3.1. Modeling Population Gowth ,P = 5.647, and = The plot of the logistic solution with these paametes is shown in the following figue P t c) If we use only thee points to compute the paametes, the fit of the logistic model is vey poo. Howeve, if we use all of the data the fit is quite good The data is shown in the following figue plotted as points. Using this we estimate K = 999. To find we use the solution to the logistic model KP Pt) = P + K P )e, t and solve fo = 1 ) P t)k t ln P ). P K Pt)) If we use t = 6, with P6) = 551, we get =.8. The logistic solution with these paametes is the solid cuve in the figue. 1 Population in thousands Time in days The fit seems to be quite good. Howeve, this method depends on the choice of middle point. If that is an outlie the fit might not be so good as it was in this case a) Since the population is measued in thousands, the fishing ate is h =.1 thousand fish pe day. The modified model is P =.1P1 P/1).1. b) We look fo oots of.1p1 P/1).1 =. This leads to the quadatic equation P 2 1P 1 =, which has oots P = 5 ± 15. Using the gaph of.1p1 P/1).1 we see that P 1 = is an unstable equilibium point, and P 2 = is an asymptotically stable equilibium point. c) We have P =.1P1 P/1).1 > fo P 1 <P <P 2 and negative elsewhee. It follows using qualitative analysis that fo any stating population geate than P 1, the population tends to the stable

6 18 Chapte 3. Modeling and Applications equilibium at P 2. Fo any stating population smalle than P 1, the population deceases until it dies out. In paticula a population of 1 is doomed while a population of 2 tends to P a) In the following figue, the five cuves ae given by initial conditions P) = 41, P) = 414, P) = 415, P) = 42, and P) = 45. Notice that the solution cuve with initial condition P) = 415 neve eaches zeo, although the cuve with initial condition P) = 414 does. Thus, the citical population is between 414 and 415. P t 1 b) The citical initial population will be the initial condition fo a solution cuve which has P = when t = 3. Witing the diffeential equation fo t<3gives dp dt =.38P.38P 2 2. Sepaate vaiables. dp.38p P 2 = dt. Integating this, one obtains t + C = Using conditions P3) =, one obtains ) 2.76P +.38 tan C = ) 2.38 tan o C =.867. Now setting t = and solving fo P = P, one obtains P = We will use days and 1 individuals as ou units. We ae given that the caying capacity is K = 1, P) = 1, and with t 1 = 2.3/24, KP 2 = Pt 1 ) =. P + K P )e t 1 Solving fo,weget = The havesting ate duing the last fou hous of each day is 15/h = 36,/day. Hence the havesting ate is {, in the fist 2 hous ht) = 36, in the last 4 hous

7 3.2. Models and the Real Wold 19 of each day. Using a squae wave function to model this in ou solve we get the following gaph of the solution. Population in thousands Time in days The population at the end of each day is appoximately 1, a) The havesting ate is.1p, so the modified model is P =.1P1 P/1).1P. b) The model equation simplifies to P =.1P9 P).This is anothe logistic equation. Thus P = is an unstable equilibium point, and P = 9 is an asymptotically stable equilibium point, epesenting the caying capacity. c) The population tends to the caying capacity P = The havesting ate is γp, so the modified model is P = P1 P/K) γp.this can be witten as P = γ)p1 P/K γ ), whee the new caying capacity is K γ = γ)k/.if γ<, this is again a logistic equation with caying capacity K γ, so all populations tend to this value. If γ>, then K γ <, and we see that P < wheneve P>. Thus any eal population tends to. The population dies out Refe to the answe to Execise 19. The units of γp ae individuals/ unit time. It epesents the numbe of individuals havested in a unit of time. Fo this eason it is called the yield. Notice that fo values of γ<the model is anothe logistic equation with caying capacity K γ = γ)k/.fo populations at the caying capacity the yield is γk γ = γ γ)k/.this function of γ is maximized when γ = /2. The caying capacity fo this value of γ is K/2 and the yield is K/4. Section 2. Models and the Real Wold 2.1. We have P = P179) = , and P18) = = P e 1. Solving the last equation fo we get = ln535925/ )/ We have h = 6, and the thee data points P = P179) = 3929, P 1 = P185) = 23192, and P 2 = P191) = Thus using 1.17) we get = 1 ) h ln P2 P 1 P ) =.313. P P 2 P 1 ) With this value we use the fist equation in 1.16) to find the caying capacity K = P P 1 1 e h ) = P P 1 e h 2.3. Answes will vay depending on the data chosen.

8 11 Chapte 3. Modeling and Applications Section 3. Pesonal Finance 3.1. a) Let Pt) epesent the balance in the account afte t yeas. Because the initial deposit at time t = is P = $12, P =.5P, P) = 12. Sepaate the vaiables and solve fo P, then use the initial condition to poduce Pt) = 12e.5t. Thus, the amount in the account afte 1 yeas is b) Solve Pt) = 5, o P1) = 12e.51) $ e.5t = 5 e 1/2)t = 25 6 t = 2 ln 25 6 t 28.5y 3.2. Let Pt)epesent the balance in the account t yeas afte the initial investment. Let P epesent the initial investment, i.e., P) = P. Consequently, P = P, P) = P. Sepaate vaiables and use the initial condition to poduce P = P e t. Because the initial investment doubles in five yeas, 2P = P e 5), 2 = e 5, = ln 2 5,.1386, o, 13.86% Let Pt) epesent the balance in the account t yeas afte the initial investment. Let epesent the annual ate, d the yealy deposit, and P the initial investment. Then, P = P + d, P) = P. This equation is linea, with integating facto e t. Consequently, e t P ) = de t, Use P) = P to poduce C = P + d/ and e t P = d e t + C, P = d + Cet. Pt) = d + P + d ) e t.

9 3.3. Pesonal Finance 111 Thus, the amount in the account afte 1 yeas is P1) = ) e.61) $25, Let Pt) epesent the balance in the account t yeas afte Jason s day of bith. Let epesent the annual ate, d the annual deposit. Since no initial investment is equied, P = P + d, P) =. This equation is linea, with integating facto e t. Consequently, e t P ) = de t, Use P) = to poduce C = d/ and e t P = d e t + C, P = d + Cet. Pt) = d e t 1 ). Because P18) = 5, 5 = d e 18) 1 ), d = 5 e 18 1, d = 5.625) e ) 1, d $1, Let Pt) epesent the balance in the account afte t yeas. Let epesent the annual ate, w the yealy withdawal, and P the amount of the inheitance. Then P = P w P) = P. The equation is linea with integating facto e t. Consequently, e t P ) = we t, e t P = w e t + C, P = w + Cet. Use P) = P to poduce C = P w/ and Pt) = w + P w ) e t. Now, to find when the funds ae depleted, set Pt) =. = w + P w ) e t, e t = w/, w/ P t = 1 ln w/ w/ P.

10 112 Chapte 3. Modeling and Applications Thus, the account will be depleted in t = 1.5 ln 8/ yeas. 8/ Let Pt)epesent the loan balance afte t yeas. Let epesent the annual ate, w the annual payment, and P the amount of the loan. Then P = P w P) = P. The equation is linea with integating facto e t. Consequently, e t P ) = we t, e t P = w e t + C, P = w + Cet. Use P) = P to poduce C = P w/ and Pt) = w + P w ) e t. Now, the loan is exhausted at the end of fou yeas. Consequently, P4) =, so = w + P w ) e 4), w e 4 1 ) = P e 4, P = w 1 e 4 ) P = 225)12) 1 e 4.8) ),.8 P $9, 242, Let Pt)epesent the loan balance afte t yeas. Let epesent the annual ate, w the annual payment, and P the amount of the loan. Then P = P w P) = P. The equation is linea with integating facto e t. Consequently, e t P ) = we t, e t P = w e t + C, P = w + Cet. Use P) = P to poduce C = P w/ and Pt) = w + P w ) e t.

11 3.3. Pesonal Finance 113 Now, the loan is exhausted at the end of 3 yeas. Consequently, P3) =, so = w + P w ) e 3), w e 3 1 ) = P e 3, w = P 1 e, 3 w =.81) 1 e 3.8) w $8, Let Pt)epesent the loan balance afte t yeas. Let epesent the annual inteest ate, w the annual payment, and P the amount of the loan. Then P = P w, P) = P. The equation is linea with integating facto e t. Consequently, e t P ) = we t, Use P) = P to poduce C = P w/ and e t P = w e t + C, P = w + Cet. Pt) = w + P w ) e t. Now, $1, pe month makes $12, pe yea, so w = 12. Futhemoe, the tem of the loan is 3 yeas, so P3) = and = w + P w ) e 3), we 3 w = P e 3, P = w e 3 1 ), e 3 P = 12 e 3.725) 1 ),.725e 3.725) P $146, Let St) epesent José s salay afte t yeas. Consequently, S =.1S, S) = 28, whee S is measued in thousands of dollas. Theefoe, José s salay is given by St) = 28e.1t. Let Pt) epesent the balance of the account afte t yeas. Let ρ epesent the fixed pecentage of José s salay that is deposited in the account on an annual basis. Thus, P =.6P + 28ρe.1t, P) = 2.5,

12 114 Chapte 3. Modeling and Applications whee Pt)is also measued in thousands of dollas. Multiply by the integating facto, e.6t, and integate. e.6t P ) = 28ρe.5t e.6t P = 28.5 ρe.5t + C P = 56ρe.1t + Ce.6t The initial condition P) = 2.5 poduces C = ρ and P = 56ρe.1t ρ)e.6t. Now, José wants $5, in the account at the end of 2 yeas, so P2) = 5 and 5 = 56ρe.12) ρ)e.62), 5 2.5e 1.2 = 56ρe.2 e 1.2 ), 5 2.5e1.2 ρ = 56e 1.2 e.2 ), ρ.35, o 3.5% Let Pt)epesent the balance afte t yeas, the annual ate, and d the annual deposit. Thus, P = P + d, P) = 1. Howeve, in this case the annual deposit is a function of time. The initial deposit is $1, and this is inceased by $5 each yea. Consequently, dt) = 1 + 5t and P = P t), P) = 1. Multiply by the integating facto, e t, and integate. e t P ) = 1 + 5t)e t e t P = P = 1 + 5t 1 + 5t 5 ) e t + C Ce t The initial condition P) = 1 gives C = 1 + 1/ + 5/ 2 and 1 + 5t P = ) e t. 2 2 At ten yeas, ) P1) = 5.6.6) ) e.61),.6) 2 P1) $46, a) Let P n) epesent the balance at the end of n compounding peiods, I the annual inteest ate, m the numbe of compounding peiods pe yea, and P the initial investment. Thus, Pn+ 1) = 1 + I ) P n), P ) = P. m b) Compae with Pn+ 1) = 1 + I ) P n), P ) = P. m an + 1) = an), a) = a,

13 3.3. Pesonal Finance 115 and note that = 1 + I/m and a = P. Consequently, an) = a n becomes P n) = P 1 + I m) n a) Let Pt) epesent the balance at the end of t yeas. Let epesent the annual ate and P the initial investment. Thus, P = P, P) = P. Consequently, Pt) = P e t, P1) = 2e.61), P1) $3, b) i) In this case, m = 2. Futhemoe, thee ae 2 compounding peiods in 1 yeas, so P2) = ) 2 $3, ii) In this case, m = 12. Thee ae 12 compounding peiods in 1 yeas, so P12) = ) 12 $3, iii) In this case, m = 365. Thee ae 365 compounding peiods in 1 yeas, so P365) = ) 365 $3, c) The fomula P = P, used to evaluate the balance when the inteest is compounded continuously, is a fai appoximation of what actually occus in the discete, eal-wold case. This appoximation is impoved as the fequency of compounding inceases in the discete case. In fact, one can show that the balance in an account is given by Pt) = P 1 + m) I mt, whee t is measued in yeas. A little calculus will eveal an inteesting identity as the numbe of compounding peiods pe yea is inceased. lim P 1 + I ) mt = P e It m + m a) The seies n 1 is geometic, finite, with sum 1 n )/1 ). Consequently, an) = a n + b n 1), 1 an) = a n n ) + b, 1 an) = a n + b 1 b 1 n, ) an) = a n + b) Compaing Pn+ 1) = b 1 b I ) P n) + d, P) = P. m

14 116 Chapte 3. Modeling and Applications with an + 1) = an) + b, a) = a shows us that = 1 + I/m, b = d, and a = P. Consequently, the solution ) an) = a n + becomes P n) = P n) = P P + md I d I/m) ) 1 + I m b 1 b 1 ) 1 + I m) n + ) n md I. d I/m), a) Let Pt) epesent the balance afte t yeas, the annual inteest ate, d the yealy deposit, and P the initial investment, Thus, P = P + d, P) = P. Multiply by the integating facto, e t, and integate. e t P ) = de t e t P = d e t + C Pt) = d + Cet The initial condition P) = P gives C = P + d/ and Pt) = d + P + d ) e t. Afte 1 yeas, P1) = ) e.51) $16, b) Let P n) epesent the balance afte n compounding peiods, I the annual inteest ate, m the numbe of compounding peiods pe yea, and P the initial investment. Then, Pn+ 1) = 1 + I ) P n) + d, P) = P, m with solution fom execise 13) P n) = P + md ) 1 + I ) n md I m I. In ten yeas, thee ae 4 compounding peiods. Thus, P4) = 2 + 4)25) ) ) 4 4)25) $16, c) One significant change is the deposit at the end of each compounding peiod at the end of each day). Because Demetios is depositing $1, pe yea, his daily deposit is d = $1/365. Next, thee ae 3,65 compounding peiods in the ten-yea peiod. Thus, P365) = )1/365) ) ) )1/365), $16,

15 3.4. Electical Cicuits a) Let Pt)epesent the balance of the loan afte t yeas, I the annual inteest ate, w the yealy payment, and P the amount of the loan. Then, P = P w, P) = P. Multiply by the integating facto, e t, and integate. e t P ) = we t e t P = w e t + C P = w + Ce t The initial condition P) = P gives C = P w/ and P = w + P w ) e t. Because the loan expies in 5 yeas, P5) = and = w + P w ) e 5), w = P 1 e, 5 w = 12).8), 1 e 5.8) w $2, , o $ pe month. b) Let P n) epesent the balance on the loan afte n compounding peiods, I the annual ate, m the numbe of compounding peiods pe yea, w the monthly payment, and P the initial amount of the loan. Then, in a manne simila to that in Execise 13, the pogess of the loan is modeled by the fist ode diffeence equation Pn+ 1) = 1 + I ) P n) w, P) = P, m with solution P n) = P mw ) 1 + I ) n + mw I m I. Thee ae 6 compounding peiods in 5 yeas, so P6) = and = P mw ) 1 + I ) 6 + mw I m I, w = P I1 + I/m) 6 m1 + I/m) 6 1), Section 4. Electical Cicuits P I w = m1 1 + I/m) 6 ), 12).8) w = /12) 6 ), w $ The model equation is Q + Q/2 = 5. The solution is Qt) = 1 1e t/ The model equation is Q + Q/2 = 5e t/1. The solution is Qt) = 25e t/1 e t/2 )/2.

16 118 Chapte 3. Modeling and Applications 4.3. The model equation is Q + Q/2 = 5 sin 2t. The solution is Qt) = 4 cos 2t + 1 sin 2t + 4e t/2 )/ The model equation is Q + Q/2 = 5 cos 3t. The solution is Qt) = 1 cos 3t + 6 sin 3t 1e t/2 )/ The model equation is Q + Q/2 = 5 t/2. The solution is Qt) = 511 e t/2 )/5 t/ The model equation is Q + Q/2 = 51 e t/1 ). The solution is Qt) = 1 25e t/1 5e t/2 )/ The model equation is I + I/1 = 1. The solution is It) = 11 e t/1 ) The model equation is I = e t/1 I/1. The solution is It) = te t/ The model equation is I + I/1 = 5 sin 2πt. The solution is It) = 5 2π cos 2πt + 2πe t/1 + sin 2πt) 1 + 4π The model equation is I + I/1 = 4 cos 3t. The solution is It) = 4 cos 3t + 12 sin 3t 4e t/1 )/ The model equation is I + I/1 = 1 2t. The solution is It) = 3 2t 3e t/ The model equation is I + I/1 = 11 e t/2 ). The solution is It) = 11 2e t/2 + e t/1 ) Qt) = EC1 e t/rc ) It) = E + RI E)e Rt/L )/R The model equation, Q + Q/2 = 1/2)e t/1 with Q) =, has solution Qt) = 25e t/1 e t/2 )/2. We find the maximum by setting the deivative [ 1 Q t) = 25 2 e t/2 1 ]/ 1 e t/1 2 =, and solving fo t. It has a maximum at t = 25 ln The maximum value is Q25 ln 5) = The model equation, I + I/1 = 1/2)e t/2 with I) =, has solution I = 1e t/2 e t/1 ). The solution is maximized at t = 2 ln with maximum value I2 ln 2) = a) b) The model equation, Q + Q/1 = 1 with Q) =, has solution Q = 11 e t/1 ) fo t 5. Note that Q5) = Q 5 = 11 e 1/2 ). Then fo the next 5 seconds, the model equation is Q = Q/1 with Q5) = Q 5. The solution to this equation is Q = Q 5 e t 5)/1.Att = 1 seconds, the amount of chage is Q = Q 5 e 1/

17 3.4. Electical Cicuits a) The nine solution ae plotted below. 1 Q t The steady state solution is a paticula solution to the inhomogeneous equation, which all solutions convege to egadless of thei initial condition. Note that this is the case fo all nine solutions coesponding to the nine diffeent initial conditions. b) The fequency of the foce is 2π/2π = 1 Hz. This also appeas to be the peiod of the steady-state esponse The nine solutions ae plotted below. 1 I t The fequency is of the foce is 2π/2 = π Hz. This also appeas to be the fequency of the steady-state esponse a) In the following plot, the solution is the solid cuve, the tansient esponse is the dotted cuve, and the steady-state solution is dashed. I.5 t 5 1.5

18 12 Chapte 3. Modeling and Applications b) In the following figue, the steady-state solution is the dotted cuve. I 1 t All solutions convege to the steady state solution as time gets lage If we use I = Q, we get the diffeential equation Q + 1 RC Q = E sin ωt. C This linea equation can be solved using the integating facto e t/rc. We get the solution EC Qt) = 1 + R 2 C 2 ω sin ωt RCω cos ωt) + 2 Ae t/rc, whee A is an abitay constant. The tem Ae t/rc dies out as t inceases, so it is the tansient tem. The fist tem does not die out and is theefoe the steady-state solution By the capacitance law, the voltage dop acoss the capacito is V = Q/C. Thus Q = CV. If we substitute this into the given equation we get the esult. The equation is linea, and it can be solved using the integating facto e t/rc. The geneal solution is E Vt)= 1 + R 2 C 2 ω RCω sin ωt + cos ωt) + 2 Ae t/rc, whee A is an abitay constant. The tem Ae t/rc dies out as t inceases, so it is the tansient tem. The fist tem does not die out and is theefoe the steady-state solution By the chain ule di dt = di dq = I di dq dt dt. Hence the equation becomes LI di + Q/C =. dq This sepaable equation has the solution I = k Q 2 /LC whee k is a constant. Since I = dq/dt, the equation now becomes dq = k Q dt 2 /LC. This is anothe sepaable equation, and dq k Q2 /LC = dt The integal on the left involves an acsine and the solution is ) t Qt) = k 1 LC sin + k 2. LC

19 3.4. Electical Cicuits 121 Since I = dq/dt, we also obtain ) t It) = k 1 cos + k 2, LC whee k 1 and k 2 ae constants.