610 Chapter 9 Further Applications of Integration

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1 6 Chapter 9 Further Applications of Integration 4. So for the orthogonals: 4. c Ê c Ê Ê Ê Ê Ê C Ê É C, C ab ˆ Ê % Ê ˆ ˆ ab Ê ˆ 5. k Ê k Ê k a bab. So for the orthogonals: Ê Ê ln C 4 Ê Ê ln / ln ln C / C kk 6. c Ê 4 Ê. For orthogonals: ln ln C Ê Ê c c e ae bab e c c ae b 7. ce Ê c Ê! Ê e e Ê. So for the orthogonals: Ê Ê C Ê C Ê È C k ln Š ln 8. e Ê ln k Ê k Ê ln Ê Š ln Ê. So for the orthogonals: ln Ê ln Ê ln 4 a b ˆ C ln C Ê

2 Section 9.5 Applications of First-Orer Differential Equations an intersect at a, b. Also, 3 5 Ê 4 6 Ê Ê a, b Ê 3 Ê Ê a, b. Since ˆ ˆ, the curves are orthogonal. 3. (a) Ê C is the general equation of the famil ith slope CÑ. For the orthogonals: Ê Ê ln ln C or C (here C e is the general equation of the orthogonals. (b) Ê Ê Ê Š Ê ln ln C Ê C is the equation for the solution famil. ln ln C Ê Ê Ê slope of orthogonals is Ê Ê C is the general equation of the orthogonals.. 4a 4a an 4b 4b Ê (at intersection) 4a 4a 4b 4b Ê a b aa bb Ê aa bbaa bb aa bb Ê a b. No, 4a 4aaa bb 4a 4a 4ab 4ab Ê Èab. Thus the intersections are at Š a b, È 4a 4a ab. So, 4a 4a Ê hich are equal to an Š Èab 4a Èa an Èa 4b 4b at the intersections. Also, 4b 4b Ê hich are equal to an ab b b Š È Š Èab É an É at the intersections. a b a b. Thus the curves are orthogonal. 4b b b Š Èab a a

3 6 Chapter 9 Further Applications of Integration CHAPTER 9 PRACTICE EXERCISES. È cos È Ê Ê tanè C Ê ˆ tan ˆ C È cos È 3a b a b 3. Ê 3a b Ê ln a b C sinˆ secab 3. seca bsec Ê sec Ê tan C Ê sina b tan C sin cos ab cosab cosab 4. cos a b sin Ê Ê C Ê É C 5. e e e C e È Ê È a b Ê 3/ a3 4b a b 3 4 Ê 3/ a b C 5 5 3/ a b a34b 3/ a b a34b 5 5 Ê ln C Ê ln C 6. e Ê e Ê ln e C cos sec 7. sec cos Ê Ê tan cos sin C 3/ 8. 3È csc Ê 3È Ê a bcos 4 sin C 3/ Ê a bcos sin C e csc 9. Ê e Ê a be ln k k C e e e e csc. e csc Ê csc Ê e Ê asin cos b a be C a b. a b Ê a b Ê Ê ln lna blnabc Ê ln ln a bln a bln C Ê ln lnš Ê Cab Cab c lnš b. a ba b Ê Ê ln C Ê lnš ln ln C Ê C / / a b ˆ / 3. e Ê e. p a b, v e e. / / / / / / / e e ˆ e ˆ ˆ e Ê ˆ e Ê e C Ê e Š C e sin Ê e sin. p a b, v a b e e. e e e e sin e sin Ê ae b e sin Ê e e asin cos bc Ê e asin cos bce 5. Ê ˆ. ln ln v a b e e e. C Ê ab Ê CÊ

4 Chapter 9 Practice Eercises ln Ê ˆ ln. vab e ln e. ˆ ˆ ln Ê ˆ ln Ê cln C Ê cln C e ec e aeb e ab e lnae b b 7. a e b ae e b Ê a e b e e Ê. v a b e e e. ae b ae bˆ e ec ae b Ê cae b e Ê ae b e C e aeb ecc ecc e e Ê 8. 4e Ê 4e. Let vab e e. Then e e 4e Ê ae b 4e Ê e a be C Ê a be Ce 3 9. a 3 b Ê 3 Ê ab 3 Ê C ln 3 ln 3. a3 cos b Ê Š cos. Let vab e e e Then 3 cos an cos sin C. So asin Cb a b a b. e Ê e e Ê e e C. We have ab, so e e C Ê C e an ab e e e Ê lnˆ a b e e ln tanc e bc tanc e bc ln tan ca b C e tancabbln. Ê Ê lnaln b tan abc Ê e. We have ab e Ê e e Ê e Ê tan abc ln Ê C ln Ê C ln Ê e a b a b a b 3. a b ˆ ln ln Ê. Let vab e b a b a b e e a b. So a b a b a b Ê a b Ê a b a b Ê a b C Ê a b Š C. We have ab Ê C. So 3 3 ab Š ln 3 4. Ê ˆ ˆ. Let vab e e. So 4 3 C a b 4 4 a b Ê Ê C Ê. We have Ê C Ê C. So Let vab e e. So e 3 e e Ê Š e e Ê e e C We have ab Ê e ab e C Ê C Ê C an e e Ê e a b sin 6. a cos b Ê cos Ê ˆ cos ln. Let vab e e. So ˆ cos Ê cos Ê cos Ê sin C. We have ˆ Ê ˆ C Ê C. So sin Ê 7. ˆ È Ê Ê lnˆ È ln C. We have ab Ê lnš È ln C ˆ È Ê ln C ln ln 4. So lnˆ ln ln 4 ln 4 Ê lnˆ È a b È lna4b lna4b /

5 64 Chapter 9 Further Applications of Integration ˆ È / a b Ê e e Ê È È Ê ˆ È ln ln e e ab e e c Ê Ê e e C. We have ab Ê e e C Ê C. 3 3 /3 3 3 So e e Ê 3ae e b Ê c3ae e b ˆ c 9. 3 e 3 e Ê. Let v e e 3 a b ˆ ln e a b. So e e e e ˆ 3 Ê ˆ 3 Ê 3 C. We have ab Ê 3abC Ê C 3 e Ê 33 Ê e a33b ln 3 3. a3 b Ê Ê Ê Š. 3 P a b Ê P a b 3ln Ê v a b e e e e Š e Ê e e e a bc 3 ˆ Ce abcec 3 ˆ 4e Ê. We have ab Ê ÊC 4e an Ê 3. To fin the approimate values let n n an cos nba. b ith,, an steps. Use a spreasheet, graphing calculator, or CAS to obtain the values in the folloing table To fin the approimate values let z aa ba 3bba. b an n n n n a ba 3baz ba 3b n n nc nc n n Š a.bith initial values 3,, an steps. Use a spreasheet, graphing calculator, or CAS to obtain the values in the folloing table

6 z n n n n nc nc n Chapter 9 Practice Eercises 65 nc nc nc nc n n 33. To estimate a3 b, let z Š a.5 b an Š a.5 b ith initial values,, an 6 steps. Use a spreasheet, graphing calculator, or CAS to obtain a3b.963. n n nc n 34. To estimate a4 b, let z c nc Š a.5 bith initial values,, an 6 steps. Use a spreasheet, graphing calculator, or CAS to obtain a4b e nc nc 35. Let n n ˆ b b a bith starting values an, an steps of. an.. Use a spreasheet, programmable calculator, or CAS to generate the folloing graphs. (a) (b) Note that e choose a small interval of -values because the -values ecrease ver rapil an our calculator cannot hanle the calculations for Ÿ. (This occurs because the analtic solution is lna e b, hich has an asmptote at ln.69. Obviousl, the Euler approimations are misleaing for Ÿ.7.) z e nc nc e nc nc ezn n n 36. Let zn n c nc n Š a ban n n c nc n n Š a bith starting values an, an steps of. an.. Use a spreasheet, programmable calculator, or CAS to generate the folloing graphs. (a) (b) Ê C ÊC Êaeactb 3 Ê a b is the eact value. Ê Ê C; an

7 66 Chapter 9 Further Applications of Integration Ê Ê lnkkc; an Ê ln C ÊC Êaeactb lnkk Ê ab ln.369 is the eact value C C Ê Ê lnkk C Ê e e e C e ; an / / / Ê Ce ÊC e aeactb e e ˆ e / 3/ Êab e is the eact value Ê Ê C; an C Ê C Ê Êaeactb È Êab È3.73 is the eact value. 4. Ê a ba b. We have Ê a b, a b Ê,. (a) Equilibrium points are (stable) an (unstable) (b) Ê Ê a b a ba b. So Ê,,. (c) 4. Ê a b. We have Ê a b Ê, Ê,. (a) The equilibrium points are an. So, is unstable an is stable. (b) Let ïî increasing, íï ecreasing.!!! qqíïïïïïqqñqqïïïïïîqqñqqíïïïïïqqp 3 3 Ê Ê a ba b Ê 3 a 3 b Ê a ba b. So, Ê,, Ê,,

8 Chapter 9 Aitional an Avance Eercises 67 (c). Let ïî concave up, íï concave on. qíïïïïïqqñqqïïïïïîqqñqqíïïïïïqqñqqïïïïïîqp!!!! / v v s v s s v Ê Ê s Ê Ê v gr gr gr gr s C v s C s C. When t, v v an s R v R C gr C v gr v s v gr È gr gr s ÈgR s s È Ès 43. (a) Force Mass times Acceleration (Netons Secon La) or F ma. Let a v. Then ma mgr s a gr s v gr s v v gr s s v v gr s s Ê Ê Ê Ê Ê (b) If v gr, then v Ê v É, since v if v gr. Then Ê Ès s ÈgR / Ê s s ÈgR 3/ Ê s ÈgR 3/ 3 t C Ê s gr 3 ˆ È t C; t an s R ÊR 3/ ˆ 3ÈgR C ÊC R 3/ Ês 3/ 3 gr t R 3/ 3 a b ˆ È ˆ RÈg t R 3/ 3/ 3 / 3/ 3 gr ˆ È È 3/ 3v 3v R R R R R g t R Š t R ˆ t Ê s R ˆ t /3 vm 44. coasting istance vm Ê a ba b.97 Ê k sab t ˆ e ak/mbt Ê sab t.97ˆ e a7.343/3.84bt k k k.8866t Ê sab t.97a e b. A graph of the moel is shon superimpose on a graph of the ata. CHAPTER 9 ADDITIONAL AND ADVANCED EXERCISES A A A A A V V c V c V V A C k Ê c e e V t. Appl the initial conition, ab Ê c C Ê C c. (a) k ac b Ê k a cb Ê k Ê k Êlnk ck k t C k A V t Ê ca cbe. A (b) Stea state solution: _ limab t lim k t ca cbe V c a cbab c tä_ tä_. Measure the amounts of ogen involve in ml. Then the inflo of ogen is ml/min (Assume: it ill take 5 minutes to eliver the 5L 5mL); the amount of ogen at t is ml; letting A the amount of ogen in the A flask, the concentration at time t is A ml/l; the outflo rate of ogen is A ml/l (lb/sec). The rate of change in A,, equals the rate of gain ( ml/min) minus rate of loss (A ml/min). Thus: A A t A Ê A Ê lnaa b t C Ê A Ce. At t, A, so C 79 an t ml A 79e. Thus, Aa5b 79e ml. The concentration is 99.47%. ml

9 68 Chapter 9 Further Applications of Integration 3. The amount of CO in the room at time t is Aab t. The rate of change in the amount of CO, is the rate of internal prouction (R ) plus the inflo rate (R ) minus the outflo rate (R 3). breaths/min R 3 stuents ft 3.4 ft 3 ˆ a bˆ Š CO.39 ft 3 CO A stuent 78 ft3 min R Š Š.4.4 ft ft CO ft CO min min min R Š.A 3 A A 3 ft CO, min.39.4.a.79.a Ê A.A.79. Let vab t e. We have Š Ae.79e....t Ê Ae.t.t 3.79e 7.9e C. At t, A a,ba.4b 4 ft CO Ê C 3.9.t.a6b 3 3 Ê A e. So Aa6b e 7.87 ft of CO in the, ft room. The percent of 7.87 CO is,.8% mv a b m mv a b m v m m m v m m b Êm kbkt C. At t, m m, so C m an m m kbkt. 4. F av ub Ê F av ub Ê F m v v u Ê F m u. v v ub k k m kbt k m kbkt m Thus, F am kbktb ukbk am kbktbkgk Ê g Ê v gt u lnš C v at t Ê C. So v gt u lnš Ê gt u lnš an u c, at m kbkt m t Ê gt c t Š lnš m kbkt m kbkt kbk m m k b k t m P a b 5. (a) Let be an function such that vab vabqab C, vab e. Then P a b P a b av a b b v a b vab vq. a b a b We have v a b e Ê vab e P a b vp. a b a b Thus vab vabpab vabqab Ê Pab Qab Ê the given is a solution. (b) If v an Q are continuous on ca, b an aa, b b, then vab t Qab t vabqab Ê vatbqatb vabqab. So C va b vabqab. From part (a), vab vabqab C. Substituting for C: vab vabqab va b vabqab Ê vab va b hen. a b 6. (a) P,. Use v P a b a b a b e as an integrating factor. Then avabb Ê vab C Ê Ce an C e, C e,, C C e a b a b a b P C e a b 3 an. So is a solution to P a b ith a b. (b) v P a b a a bc a b a bb Š e P a b e ac C b ac C b ac b!. P a b P a b P a b P a b 3 v v a a bc a b a bb a a bc a b a bb! C P a b P a b P a b P a b a b (c) C e, C e,. So Ê C e C e! Ê C C Ê C C Ê ab a b for a b.