Problem Set 7-7. dv V ln V = kt + C. 20. Assume that df/dt still equals = F RF. df dr = =

Transcription

1 20. Assume ha df/d sill equals = F RF. df dr df/ d F RF = = 2 dr/ d R 0. 04RF 0. 01R 10 df R= 70 and F = 1 = = K dr F (70, 30) (70, 1) R 100 Noe ha he slope a (70, 1) is abou The fo and rabbi populaions spiral oward a fied poin. Again, and even more surprisingl, he rabbis sabilize a R = 40 (4000). Bu he sable fo populaion is reduced o or. Along he wa, he model shows ha he foes are reduced o abou 1, hus becoming in danger of eincion! 23. See he graph in Problem 21 wih iniial condiion (70, 30). Wih his man foes and huners chasing rabbis, he rabbis become einc. A his poin, he foes have been reduced o jus. Afer he rabbis become einc, he foes decrease eponeniall wih ime, evenuall becoming einc hemselves. Problem Se 7-7 Review Problems R0. Answers will var. R1. P() = 3(0. ) P () = 3(0. ) ln 0. P() P () P ()/P() P () 3( 0. )ln 0. = = ln 0. P () 3( 0. ) = , which is a consan, Q.E.D. R2. a. V = speed in mi/h; = ime in s dv = V d b. dv = d V ln V = + C + C C V = e = e e V = C 1 e C 1 can be posiive or negaive, so he absolue value sign is no needed for V. In he real world, V is posiive, which also maes he absolue value sign unnecessar. c. 400 = Ce 0 C = = 400e 40 ln 1. 2 = = V = 400e d. 70 = 400e ln 1. 7 = = s / 2 R3. a. = 6 = ( 3+ C) b. = (3 4) 2 ( = (3 14) 2 does no wor because a (3, ), / = 30 bu 6 1/2 = 30.) c. 10 (3, 2) d. A = 2, = 12 and = 4. See graph in par c. A line hrough (2, 4) wih slope 12 is angen o he graph. e. i. dn/d = 100 N dn = d 100 N (1/) ln 100 N = + C Using (0, 0) gives (1/) ln 100 = C. Subsiuing his value for C gives (1/) ln 100 N = (1/) ln 100. ln 100 N ln 100 = ln 1 (/100)N = 1 ( / 100) N = e N = ( 100 / )( 1 e ) Using (7, 600) and solving numericall gives N = (1 e ) ii. = 30: Abou 42 names 0 Problem Se 7-7 Calculus Soluions Manual 200 Ke Curriculum Press

2 iii. lim N = (1 0) = The brain sauraes a abou 2211 names. iv. Le dn/d = = 100 N N = = names. Subsiuing his for N gives K= K( 1 e ) e = K = 03. (eacl) K ln 03. = = das K or: 30 = N() N( 1) ( 1) = K[ e + e ] ( 1) = Ke ( e + 1) 27 das 20 R4. a. = A (2, ), / = 1.7. A (10, ), / = The slopes a (2, ) and (10, ) agree wih hese numbers. b. Iniial condiions (1, ) and (1, 12) (1, 12) (10, ) Table wih iniial condiion (1, ), = 1: ( = 1) ( = 0.1) M M (1, 10) (1, ) (2, ) (1, ) R. a. The soluion conaining (1, ) crosses he -ais near = 7, converges asmpoicall o he -ais as approaches zero, and is smmeric across he -ais. The soluion conaining (1, 12) goes o infini as goes o infini. c. See he graph in par b wih iniial condiion (1, 10). The soluion conaining (1, 10) behaves more lie he one conaining (1, 12), alhough a sligh discrepanc in ploing ma mae i seem o go he oher wa. 20 = = 0.1 = 1 For = 1, he graph crosses he -ais a abou = 11. b. See he able in par a for = 0.1. See he graph in par a. c. The accurac far awa from he iniial condiion is ver sensiive o he size of he incremen. For insance, in par a he firs sep aes he graph so far down ha i crosses he -ais before running off he edge of he grid. The greaer accurac wih = 0.1 shows ha he graph acuall does no cross he -ais before = 20. Calculus Soluions Manual Problem Se Ke Curriculum Press

3 R6. a. d. Coninuing he compuaions in par c, he graph crosses he -ais close o = 2.. See he able in par a. 10 (hundred beavers) (Noe ha he general soluion o he differenial equaion is ( 6) 2 + 2( 7) 2 = C, and he specific soluion for he given iniial condiion is ( 6) 2 + 2( 7) 2 = 0, whose graph is a single poin.) e. Iniial condiion (, 7) (6, 7) (, 7) (1, 7) (1, 7) 10 (ears) The populaion is decreasing because i is above he maimum susainable, 00 beavers ( = ). B Euler s mehod,.3, or abou 36 beavers, a = 3 ears. b. See he graph in par a wih iniial condiion (3, 100), showing ha he populaion is epeced o increase slowl, hen more rapidl, hen more slowl again, leveling off asmpoicall oward 00. This happens because he iniial populaion of 100 is below he maimum susainable. c. = 06. = ae Subsiue ino he general equaion. Subsiue he iniial 1 = ae. condiion (3, 1). a = e 1. = Solve for a. = + = 1 e 1. e e 06.. K The poin of inflecion is halfwa beween he asmpoes a = 0 and =. 4. = 1 + e e Subsiue 4. for. = ln (e 1. )/0.6 = r d. =.( 0 6 ) ( 7) = 0 when = 6, and = 0 when = 7. So he sable poin is (6, 7), corresponding o he presen populaion of 600 Xalos naives and 7000 as. Suddenl here are oo man predaors for he number of pre, so he a populaion declines. Because is decreasing from (, 7), he graph follows a clocwise pah. f. See he graph in par e wih iniial condiion (1, 7). The graph crosses he -ais a 14.4, indicaing ha he as are huned o eincion. (The Xalos would hen sarve or become vegearian!) g. See he graph in par e wih iniial condiion (1, 7). The graph never crosses he -ais, bu crosses he -ais a 2.3, indicaing ha he a populaion becomes so sparse ha he predaors become einc. (The a populaion would hen eplode!) Concep Problems C1. a. 12 = / = 2 2 = + C, so = [ 0. ( + C)]. b. The differenial equaion would have o become 13 / afer i is inegraed. So he original equaion would have o conain 23 / afer he variables have been separaed. Conjecure: 23 / = c. Confirmaion: 23 = / 23 / = 3 1/ 3 = + C = [(1/3)( + C)] 3, a cubic funcion, Q.E.D. d. For n 0, ( n 1) / = n ( n 1) / n 1/ n = n = + C = [(1/n)( + C)] n 2 Problem Se 7-7 Calculus Soluions Manual 200 Ke Curriculum Press

4 C2. a. b. For eample: 7 / 7 / = = / = + C = [( 1/ )( + C)] Tice Price People N 00 P 6 Limi is 10, indicaing maimum possible populaion. b. a = , c = , and = , eiher b wice aing logarihms as suggesed, or b his mehod: Taing ln once ln a ce = ln P, so ln a ce 10 = ln 17 ln a c = ln 203 ln a ce = ln 226 Then subsiuing ln a = c + ln 203 ino he firs and hird equaions gives c(1 e 10 ) = ln 17 ln 203 c( 1 e ) = ln 226 ln Subsiuing c( 1 e ) = c( e 1) e = e ( ln17 ln 203) ino he previous equaion ields ln 226 ln 203 ln 226 ln 203 e = = ln 17 ln 203 ln 203 ln 17 1 ln 226 ln 203 so = ln = 0. 01K. 10 ln 203 ln 17 Then find c using c( 1 e ) = ln 226 ln 203 and find a using 203 = ae c. c = and a = g() Funcion behaves (more or less) linearl. Le N = number of ices and P = number of $/ice. B linear regression, N 0.3P , wih correlaion coefficien r = c. Le M = oal number of dollars. M P N = P( 0.3P ) M 0.3P P d. Maimize M: M 11.66P M = 0 P = Maimum M a P because M changes from posiive o negaive here (or because he graph of M is a parabola opening downward). Charge $3.30 or $3.3. e. M has a local maimum a his price because charging more han he opimum price reduces aendance enough o reduce he oal amoun made, whereas charging less han he opimum price increases aendance, bu no enough o mae up for he lower price per ice e C3. a. g () = 10e The graph does loo lie Figure 7-7e e lim e 0 lim 10e = 10e 00 e. = 10 = Noe ha his model predics an ulimae populaion of lim P ( ) 421 million. c. Now a = 1., c = 0.21, = , and he ulimae populaion is lim P ( ) 1 million. Thus, he Gomperz model is quie sensiive o a small change in iniial condiions. The prediced ulimae populaion increased b 130 million wih onl a 1 million change in one daa poin! C4. dv/d = 2V 1/ 2 + F, where F is a consan. dv = d F 2V The inegral on he righ is no he inegral of he reciprocal funcion because he numeraor canno be made he differenial of he denominaor. A slope field gives informaion abou he soluions. The following graph is for F = 20 f 3 /min flowing in. (The dashed line shows he soluion wih F = 0, he original condiion.) Saring wih f 3 in he ub, he volume levels off near 100 f 3. Saring below 100 f 3, he volume would increase oward 100. Calculus Soluions Manual Problem Se Ke Curriculum Press

5 V F = 20 F = 0 14 If he inflow rae is oo high, he ub will overflow. The ne graph is for F = 40 f 3 /min. In his case, he sable volume is above he iniial f 3. V F = 40 F = 0 14 I is possible o anidiffereniae he lef side b he algebraic subsiuion mehod of Problem Se -11, Problems The general soluion is F + C= ln ( F 2V ) V 2 and he paricular soluion for V = a = 0 is F F 2 14 = ln V 2 F 2V Unforunael, i is difficul or impossible o solve for V. The volume will asmpoicall approach F 2 /4, overflowing he ub if F 2 /4 > ub capaci. Chaper Tes T1. = T2. Solving a differenial equaion means finding he equaion of he funcion whose derivaive appears in he differenial equaion. T3. The general soluion involves an arbirar consan of inegraion, C. A paricular soluion has C evaluaed a a given iniial condiion. T4. (0, 4) T. The concave side of he graph is up, so he acual graph curves up from he Euler s angen lines, maing he Euler s mehod values an underesimae. (Or: The conve side of he graph is down, so he Euler s angen lines are below he acual graph.) T6. General logisic differenial equaion: M = M T7. = 04. = 04. ln = C = e C e 0.4 = C 1 e 0.4 = C 1 e 0.4(0) = C 1 = e 0.4 T. = 12 = 12 2 = 12+ C dp T. a. = P P = Ce d P = 3000 a = 0 P = 3000e b. P = 2300 a = = ln = K 3000 P(2) = 74.6 Phoebe will no quie mae i because he pressure has dropped jus below 00 psi b ime = 2. or: 00 = e 0314 K 1 00 = ln = 24. 7K K 3000 Phoebe will no quie mae i because he pressure has dropped o 00 jus before = 2. T10. a. = number of grams of chlorine dissolved = number of hours since chlorinaor was sared = 30 d = d 30 1 ln 30 = + C ln 30 = + C 1 30 = C 2 e = 0 when = 0 C 2 = 30 = 30(1 e ) e = 30 ( 1 ) The rae of escape is = 13 when = 100. So = = e = e ( ). K( 13 ). 4 Problem Se 7-7 Calculus Soluions Manual 200 Ke Curriculum Press

6 0. 13 b. 200 = ( 1 e ) e = 1 = K K ln K = = 1. 4K 1. hr T11. a. = 0. = = a = 7 1 ae ae = + 1 7e. 0 b. A = 0, = 2: = 0.(2)( 2)/()(0.1) = 0.07 A = 0.1, = 2.07, so = 0.(2.07)( 2.07)/()(0.1) = A = 0.2, = The precise soluion is = = 1 + 7e , which is greaer han 2.172, as epeced because he graph is concave up (conve side downward). c. 4 = e. 0 = [ln (3/7)]/ 0. = 1.64 Abou 1 monh 21 das d. T12. a. (hundred lilies) (monhs) The graph shows ha he number of lilies is epeced o decrease oward 00 ( = ) because of overcrowding. The graph sars going downward and o he righ from (0, 700) because he cooe populaion is relaivel high, hus decreasing he number of roadrunners. b. There can be wo differen values for he roadrunner populaion for a paricular cooe populaion because he wo evens happen a wo differen imes. For eample, cooes are increasing from 0 when here are 700 roadrunners, bu laer he are decreasing from 0 when here are abou 200 roadrunners. T13. Answers will var. Problem Se 7- Cumulaive Review, Chapers v() 200 (, v()) v() d represens he disance raveled in ime d. 2. Definie inegral ( ) d = = 120 mi 4. M 100 = M 1000 = The Riemann sums seem o be approaching 120 as n increases. Thus, he 120 ha was found b purel algebraic mehods seems o give he correc value of he limi of he Riemann sum. v() 200 (, v()) 0 R (roadrunners) (0, 700) 6. An Riemann sum is bounded b he corresponding lower and upper sums. Tha is, L n R n U n. C (cooes) B he definiion of inegrabili, he limis of L n and U n are equal o each oher and o he definie inegral. B he squeeze heorem, hen, he limi of R n is also equal o he definie inegral. Calculus Soluions Manual Problem Se Ke Curriculum Press