Solutions to the Exercises of Chapter 11

Transcription

1 A. Radioactive Deca Solutions to the Exercises of Chapter. First use a calculator to show that ln , ln , and ln Then check that the points (0, 0), (40, ln 0.60), (80, ln 0.36),and (0, ln 0.) do fall on a z t straight line shown. The negative slope λ of this line can be computed b using (0, 0) and each of the other points. So λ ln , λ ln , λ ln The conclusion λ 0.08 is certainl reasonable. The half-life of radium-0 is h What are the units? Seconds!. minutes 300 seconds. So there are (300) 0 e 0.08(300) atoms left. So (300) 0.0 or.%. 0 Taking 0 0 9,we get 0 9 e 0.08t. Taking t,we see that ()0 9 e So atoms decaed during the first two seconds. The decaed at an average rate of atoms/second. The rate of deca at t seconds was (). Because (t) 0 9 e 0.08t ( 0.08), () ( )(0.97) Take a calculator and check that ln , ln , ln 0.9., ln , ln , ln The graphs below show the required points as well as the line determined b (0, 0) and (3,.0).

2 z t z t Because λ is the slope of this line,rutherford deduced that So λ 0.69 das. λ i. The equation 0 e 0.8t is based on λ 0.8 das.soh das. ii ,implies that e 0.8t. So e 0.8t Hence 0.8t ln and hence t das. iii e 0.8t implies that 0.8t ln.099. So t 6. das. 3. i. Using the equation λ , we get λ h 38 das ii. B Avogadro s number,30 milligrams contain ( ) atoms iii. ( )e 0.00t. iv. 4 weeks is 8 das,so (8) ( )e atoms. 6. i. The deca constant is λ So )t secs 7e ( milligrams. ii. A reformulation of the question is: (?) 7. Set )t 0 7e ( , to get 0 e ( )t,and then )t 0 e( So ( )t ln 0,and hence t ln seconds. 7. Because refers to the number of atoms that deca in one second,we start b converting gram to a number of atoms. B Avogadro s number this is ( ) If is the number of radium-6 atoms at an time t,then if we consider the measurement to have been made at t 0,we see that (0).66 0 and (0) Appling equation (b) with t 0 we get λ(.66 0 ). So λ Therefore,the half-life is h.66 0 secs sec. Converting to ears,with ear seconds,we get h secs ear secs ears.

3 Compare this with the results λ.37 0 and h 600 ears on pages secs 8. Take 8 a.m. as time t 0,and let be the number of atoms in the sample at an time t 0 in hours. What do we know? We know that (0) 300 and (9) 900, both in atoms per minute. B Section.B,we know that ln (9) λ 9, where λ is the (0) disintegration constant of the substance. So λ 900 ln (.7) hours. Hence h hours. Be aware that we mixed the units minutes and hours units 0.4 in this problem. This did not bring about an error because min canceled in (9). (0) 9. The strateg is the same as that in the problem above. Let t 0 be the moment the first observation is made. Let be the number of atoms in the sample being tested at an time t 0 in minutes. We know that (0) and (6) So 6λ ln (6) ln ln So λ (.43) 0.4 (0) mins. Therefore the half-life of the substance is h minutes Let be the number of atoms in the sample at an time t 0 in hours,where t 0 is the instant the measurement is taken. So (0) B equation (b), (0) λ(0), where λ is the disintegration constant. If we can determine (0),we will be able to solve for λ. Since we know (0) in milligrams we need onl to convert to atoms. B Avogadro s number, (0) ( 0 3 )( ) So λ (0) (0) hours.. Because 0.0% corresponds to 0.000,there are (0.48)(0.000).76 0 grams of 40 9K in one liter of seawater. B Avogadro s number this corresponds to (.76 0 ) 40 ( ) atoms. Take this measurement as having been made at time t 0. So(0) The rate at which these atoms are disintegrating is (0). B formula (b), (0) λ(0),where λ is the disintegration constant. Once this is determined in sec,a simple plug in will determine (0). Because λ , ears λ ear ear sec.69 0 sec. So (0) ( )( )4.6 atoms/sec.. i. Because 0.3% corresponds to the fraction 0.003,the mass of the potassium in the person is (0.003)(60,000) 0 grams. 3

4 ii. Radioactive potassium 40 9K constitutes 0.0% or the fraction of all naturall occuring potassium. So the mount of 40 9K in the person is (0.000)(0) 0.0 grams. iii. The estimate for the number of 40 9K atoms in the person is provided b Avogadro s number and is (0.0) ( ) iv. The deca constant is λ ears v. Taking the time of the measurement as t 0,we get (0) λ(0),where (see formula b) (0) is the rate of disintegrations and (0) is the number of atoms at that time. It remains to convert λ into sec and to compute (0). Because ear seconds, λ ear sec. So ear sec (0) ( )( ) This is the number of 40 9K atoms that disintegrate per second. B. Matter and Energ 3. Taking the 9 protons and neutrons of a 3 9U nucleus separatel,we get a mass of 9(.0076) + 43(.0090) The difference between the total mass of the individual particles in the nucleus and the nucleus as a whole is amu. Because amu corresponds (see Exercises B) to 933 MeV and joules,we see that the binding energ of a 3 9U nucleus is (.8687)(933) 743. MeV and hence (.867)( ) joules. C. Nuclear Fission. The mass before the reaction is amu. After the reaction the mass is (.009) amu. B Exercises B, amu corresponds to 933 MeV. So 0.3 amu corresponds to (0.3)(933) 08 MeV. D. Chain Reactions 6. Using Avogadro s number,we see that gram of pure uranium-3 has 3 ( ).6 0 atoms. So pound of uranium-3 has (.6 0 )(43.6) atoms. Hence if all the atoms in one pound of uranium-3 were to undergo fission, (00)( ) MeV of energ would be produced. B Exercises B,this is equal to ( )( ) joules. 4

5 7. Note that re m-total ren + r EN + + r m+ EN. So re m-total E m-total r m+ EN EN (r m+ )EN. Hence (r )E m-total (r m+ )EN and therefore, r m+ E m-total EN. r 8. During the first 0 8 seconds,the chain reaction will undergo approximatel steps. We will therefore take m 0 4 in the formula. So the energ produced is approximatel.00 0,00 (00)(00) (4.6 0 )MeV MeV (4.6 0 ) B Exercises B,this is equal to ( )( ).9 0 joules. B Exercise 6, joules corresponds to the energ in. million pounds of coal. So the energ produced b the chain reaction corresponds to the energ in pounds of coal. ( ) ( )( ) Because h ears,the deca constant is λ in (ears). Let t 0be time in ears. Taking t 0 to be now,and letting be the amount of strontium-90 in the sample in milligrams at an time t,we get 0 e λt 0e 0.077t. So () 3. milligrams. Solving 0e 0.077t for t,we get e 0.077t 4 hence 0.077t ln So t ears and E. Critical Mass 0. Let n e be the number of neutrons that escape and let n f be the number of neutrons that can produce a fission. Then there are constants a and b such that n e a(4πr ) and n f b( 4 3 πr3 ). So n e a 4πR n f b 4 a 3 πr3 b R 3a b R. 3 Therefore, ne n f is proportional to. R G. About the Moon

6 . The formula that applies is t ( )ln z(t) + where z(t) to rubidium-87 in the sample. Taking t ,we get z(t) ln + So z(t) e (. The formula that applies is t ( )ln 9.07z(t) (z) + to potassium-40 in the sample. Taking t ,we get 9.07z(t) ln + ) where z(t) is the ratio of strontium-87 is the ratio of argon-40 So 9.07z(t) e and z(t) H. Geolog and Anthropolog 3. At the rate of inch per ear,the Red Sea will separate b inches ft. inches mile 790 miles 80 ft. in 00 million ears. So its width would be between 890 and 990 miles. 4. The appropriate formula is t ( )ln z(t) + where z(t) is the ratio of strontium-87 atoms to rubidium-87 atoms in the sample after time t. So 4.67 t ( )ln 30 + ( )ln (.03) billion ears.. The operative formula is t ( )ln z(t) + where z(t) is the ratio of strontium-87 atoms to rubidium-87 atoms in the sample after time t. Because z(t) ,we get 40 t ( ) ln(.03) ears. 6. The formula that applies is t ( )ln 9.07z(t) + where z(t) is the ratio of argon-40 atoms to potassium-40 atoms in the sample after time t. Because z(t) , t ( ) ln( ) ( )( ) ears. So the mineral grain formed.6 million ears ago. 6

7 7. The operative formula is t ( )ln 9.07z(t) + where z(t) is the ratio of argon-40 atoms to potassium-40 atoms in the sample after time t. Starting with the given inequalit z(t) and using the formula as a guide,we will work toward t. So z(t) z(t) z(t) ln z(t) ( )ln It follows that t is between.88 and 3. million ears old. 8. We are given that the ratio of radioactive carbon-4 atoms to stable atoms is k.73 0 So t ( k.73 0 )ln r 0 ( )ln ( )ln ears. 9. i. r 0 grams ii. B Avogadro s number there are ( ) 4 6C atoms in one gram of carbon-4. So ( ) 60 (6.463)(4) atoms. iii. Observe that (0).3 atoms per minute. So.3 λ( ) and hence λ min. iv. h λ minutes ears This problem calls for the use of the formula t ( k )ln r 0. We are told that t is between 4.3 thousand and thousand ears and asked to sa something about. k Proceeding as suggested b the formula,we get ( k )ln r

8 k.73 ln r e.73 k r 0 e k r r 0 r 0.64 k r k ( 3. The applicable formula is t ( )ln k we get, ( ln r 0 ) k k k r 0 k r e ). Taking t and solving for r ( )( ) This number is as close to 0 as an of ou will ever encounter. So effectivel all the 4 6C atoms will have disintegrated. I. Integrals and Equations involving Derivatives 3. i. Set x(x ) C x + D x and solve for C and D. B taking common denonimators, x(x ) C x + D x C(x ) + Dx x(x ) (C + D)x C x(x ). A comparison of numerators tells us that C + D O and C. SoC and D +. Therefore, +. So x(x ) x x dx x(x ) x dx + x dx. B using the formula d ln g(x) g (x) in Section 0.3,we get dx g(x) dx ln x + ln(x ) + C. x(x ) 8

9 ii. Again,set (x )(x 3) C + D x x 3 and solve for C and D. Doing so,we get (x )(x 3) C x + D x 3 C(x 3) + D(x ) (x )(x 3) (C + D)x 3C D (x )(x 3). B comparing numerators, C+D 0 and 3C D.SoD C and 3C+C. So. C and D. Hence +. So (x )(x 3) x x 3 (x )(x 3) dx x dx + x 3 dx. B using the formula d ln g(x) g (x), we get dx g(x) iii. Set x (x )(x+3) (x )(x 3) C + D x x+3 dx ln(x ) + ln(x 3) + C. and solve for C and D. Doing this,we get x (x )(x +3) C x + D C(x +3)+D(x ) x +3 (x )(x +3) (C + D)x +3C D. (x )(x +3) Comparing numerators tells us that C + D and 3C D 0. SoC 3 D and hence 3 D + D. So 3 D and D 3. Hence C 3. Therefore, x (x )(x+3) x + 3 x+3. So x (x )(x +3) dx iv. Doing the same thing once more,we see that 3 x dx + x +3 dx ln(x ) + 3 ln(x +3)+C. x + (x + )(x 3) C x + + D C(x 3) + D(x +) x 3 (x + )(x 3) (C + D)x 3C +D. (x + )(x 3) So C + D and 3C +D. Because D C,we get 3C + ( C) ; hence 3C C and C.SoD 4. Therefore, x + (x + )(x 3) dx 4 x + dx + x 3 dx ln(x +)+4 ln(x 3) + C. 9

10 33. i. B separating variables we get d ( )(+4) 3dt. So d ( )( +4) 3dt. To solve the integral on the left,put C + D. Taking common denominators,we ( )(+4) +4 get C + D +4 C( +4)+D( ) ( )( +4) (C + D) +4C D. ( )( +4) It follows b comparing coefficients that C + D 0 and 4C D. SoD C and 4C +C. Hence C and D. We have shown that 6 6 ( )( +4) Because ln( ) and ln( + 4) are anti-derivatives of and +4 d ( )( +4) 6 d 6 d +4 6 ln( ) 6 ln( +4)+C. respectivel,we get Because 3 dt 3t+C,we get,after setting C C C,that [ln( ) ln( + 4)] 6 3t ( + C. ) We need to solve for in terms of t. B basic properties of the log function ln 8t +6C and e8t+6c Ae 8t,where A is the constant A e 6C. Using the fact that 4 when t 0,we get Ae0 A. SoA,and hence e8t. Solving for we get 4 e8t ( +4) 4 e8t + e 8t. Therefore, 4 e8t +e 8t,( 4 e8t )+e 8t, and finall, +e8t. So 4 e8t 8+4e8t 4 e 8t. ii. After separating variables,we get ( +) ( )( +4) d dt. 0

11 Proceeding as in Section.B,we get + ( )( +4) C D +4 C( +4)+D( ) ( )( +4) (C + D) +4C D. ( )( +4) B comparing numerators,we see that C + D and 4C D. To solve for C and D,add C +D to 4C D to get 6C 3. SoC and D. Therefore, + ( )( +4) So + ( )( +4) d d + +4 d ln( ) + ln( +4)+C. Because dt t + C, we get ln( ) + ln( +4)t + C C. So ln( ) + ln( +4)4t + C, where C (C C ). Using properties of the natural log and solving for we get ln [( )( +4)] 4t + C, and hence ( )( +4) e 4t+C e C e 4t A. Because 3 when t 0,we see that (3 )(3 + 4) Ae 0 A. So A 7,and hence ( )( +4)7e 4t. How do we solve this for? B using the quadratic formula! Because we see that + (7e 4t + 8) 0. Hence ( )( +4) + 8, ± 4 + 4(7e 4t +8) So the requirements are satisfied b both ± +(7e 4t +8) ± 9+7e 4t e 4t and 9+7e 4t. iii. B separating variables, d ( )( +4) t dt.

12 The integral on the left was alread solved in (i) as d ( )( +4) 6 ln( ) 6 ln( +4)+C. The one on the right is equal to t + C and hence 6 ln( ) 6 ln( +4) t + C C. Letting C C C and solving for b using properties of the log gives us 6 ln +4 t + C and ln +4 3t +6C, and hence +4 +6C e 6C e e3t 3t Ae 3t. Because (0) 6,we find that Ae0 A. So A Therefore, 0 +4 (0.4)e 3t, so (0.4)e 3t ( +4), and hence 0.4e 3t +.6e 3t. Therefore, ( 0.4e 3t )+.6e 3t and +.6e3t. It follows that 0.4e 3t is the required function. +.6e3t 0.4e 3t iv. Separation of the variables gives us + ( )( +4) d t dt. The integral on the left was alread solved in (ii) as + ( )( +4) d ln( ) + ln( +4)+C. Because the one on the right is equal to t + C,we get ln( ) + ln( +4) t + C C. Solving for and rewriting the constants,we get ln [( )( + 4)] t + C C ln [( )( +4)] t +(C C ) ( )( +4) e t +C e C e t Ae t. Using the fact that (0) 6,we get (6 )(6 + 4) Ae 0 A. So A 40,and ( )( +4)40e t. Solving for as in (ii),we get + 840e t and hence + (40e t +8) 0. It follows b the quadratic formula that

13 ± 4 + 4(40e t +8) So the requirements are met b both + ± + (40e t +8) 40e t + 9 and ± 40e t e t +9. v. B separating variables, d ( ) dt. Notice that x is an antiderivative of x. Check in the same wa that ( ) x is an antiderivative of. Therefore, ( ) d ( ) ( ) + C. Because dt t + C,we get ( ) t + C C t + C. So, t+c and therefore,. Using the condition (0),we get. So t+c C,and C. Therefore, C J. The Logistics Model t i. A look at equation (j) shows that µ 0.08 and k 0.0. So M equation (n), 4 +( 4. )e 0.08t ii. B equation (j), µ 0.0 and k So M ( 0.. ) e 0.0t e 0.08t. 0.. B (n), e 0.0t. 4. B 3. From we get µ 0.06 and k So M µ k billion. Taking t 0 in 96 we have Substituting these constants into equation (n),we get 4. + ( 4. ) e t e 0.06t. For the ear 000,we get a population of (3) 6.0 billion. For the ear 00, (4) 7.4 billion. For 00, () 8.08 billion. For 00, (8) 0.64 billion,and for the ear 09, we get (30) 3.04 billion. 3

14 36. The data fall close to the line through (8.43,0.0) and (36.,0.043). This line s slope is So an equation for the line is z ( 8.43) z or z Hence k and µ 0.0, so M t 0 correspond to 90,we get So ( ) e t The limit on the population is million e 0.0t Letting 37. i. The data do fall close to a line,namel the line through (8.4,0.0) and (.09,0.0). z The line s slope is ,and its equation is z ( 8.4) or z Hence k and µ ii. M iii. Take t 0 in 97,so Subsitituting M 39.0, 0 8.4, and µ 0.03 into equation (n),we get ( ) e 0.03t e 0.03t.

15 iv. The ear 00 corresponds to t 4,and (4) million. 38. i ii. The table provides the plot z iii. The line determined b the points (.87,0.03) and (9.3,0.08) seems to fit the data fairl well. The slope of this line is Since (.87,0.03) is on the line,we get the equation z (.87). So z Therefore, µ and k The predicted limiting value of the population is M µk million iv. Plugging ,µ0.038 and M 80.9 into equation (n) we get ( ) e 0.038t e 0.038t. v. Note that 930 corresponds to t 40, 90 to t 60, 970 to t 80, and 990 to t 00. The U.S. population data (in millions) for these ears are 3.0,.33,03.30 and 48.7,and this corresponds respectivel to the values (40) 48.7,(60) 64.04, (80) 7.63, and (00) vi The corresponding plot with z is

16 z The case for the assertion that these points fall along a straight line is not completel convincing. 39. We will assume that the court of 4000 was observed at time t 0. So 4000e µt. Because e µ,we get e µ 3.Soµ ln Hence µ Finall, d hours. µ 40. Starting with 0 e µt,we get 000e µt and () e µ. i. We get e µ 4,so µ ln4.386 and µ Therefore 000e 0.693t. ii. () 000e (0.693) 3, 000. iii. Setting 30, 000,we get e 0.693t 30. Hence 0.693t ln ,and t 4.9 hours. 4. Using 0 e µt,we get 0,00e µt and () 3,000 0,00 e µ. i. So e µ 3, Hence µ ln and µ Therefore, 0,00 0, 00e 0.39t. ii. (6) 0, 00e (0.39)6 0,000. iii. Setting 30,000, we get e 0.39t 30, and hence 0.39t.6. So 0,00 t 6.4 hours. 4. The simulation 0 e µt applies. Setting (), 000 and (7) 6, 000,we get, e µ and 6, e 7µ. i. We need to solve for 0. Because e µ 000 0,we have e µ 000. Therefore, ( 0 ) Therefore 0 6, ,000 6,000(e µ ) 7 6,000. e 7µ 000 7,hence , million. So ii. Since e µ,we get e µ So µ ln and µ So 03 e 0.79t. iii. Because µ , we get d 0.88 hours or about 3 minutes. d

17 M 43. As in Section.6B,we will start with the equation and use the data of +( M 0 )e µt the experiment to determine the parameters 0,M, and µ. A look at Table.6 tells us that 0 64, 0 and that M can taken to be 00,000,000. So , 0e. µt Using the cell counts at t 0 and t 3, we get m log 0 (3) log 0 (0) 3 0 log 0 6, 0, 000 log 0 64, Therefore µ (.3)(0.8).8 as in Section.6B. Hence , 0e..8t Substituting t,,,...,7, we get ( ) 8, 000; () 388, 000; (), 330, 000; (3) 3, 300, 000; (4) 60, 00, 000; () 44, 00, 000; (6) 88, 000, 000; (7) 98, 000, Compare Gause s equation with equation (n). Notice that if Gause s equation is rewritten as 37 +e.69 e.309t then it has the same form as (n) with M 37,µ.309, and M 0 e So M 0 77, and 0 M 77. The fact that 0 and not is some indication of the discrepanc between the mathematical model and the observed realit. 4. Consider the function f() (aµ + b aµ ) where is the number of microbes per milliliter M of medium. Note that f () aµ M + aµ + b So f () 0 when aµ aµ + b, so when M M aµ (aµ + b) M ( + b aµ ). When > M aµ (aµ+b), then >aµ+band it follows that f () < 0. When < M (aµ+b), aµ M aµ then aµ <aµ+ b and f () > 0. So f() attains its maximum value when M (aµ + b). M aµ 7

18 L. Gompertz s Model ( 46. i. Because d ln dt (t) and ) d dt me kt mke kt, we see that the functions ln and me kt have the same derivative. The therefore differ b a constant. (See Section.B for instance.) ii. From ln me kt +C, we get e ln e me kt +C e C e me kt. B taking t 0, we get 0 (0) e C e m. So e C 0 e m and hence 0 e m e me kt 0 e m me kt. iii. When t is pushed to infinit, e kt goes to zero. So e kt 0 e m e me kt goes to 0 e m. iv. Since (t) mke kt,we get (t)mke kt mk e kt mke kt mke kt mk e kt mk e kt [ me kt ]. v. After setting 0 we get me kt and hence e kt m. So t ln m. Fort> ln m k k we get kt > ln m. Because e x is an increasing function, e kt >e ln m m and hence >e kt m. It follows from (iv) that is concave down for t> ln m. In the same k wa, is concave up for t < ln m. Therefore has an inflection point when k t ln m. When m, ln m 0. So is concave up for t 0. But b assumption k k t 0. So is never concave up; hence onl concave down; and there is no point of inflection. 47. i. Using 0 80 and (see Exercise 46 iii) 00, 000, e m 80e m, we get m ln 00,000, So 80e e kt. With t 3, we get 6, e e 3k. So e 3k ln 6, Hence 3k ln and k 0.4. ii. From (i) we see that 80e t. For t,, and 6 we get the values 4,000, 4,900,000 and 84,000,000. A look at Table.7 shows that the fit of the logistics model is considerabl better. 48. i. We are given that (t) mke kt for some constants m and k and t 0. For t 0,we get mk (0) (0) 0.0. So (t) 0.0e kt. So when t 30,0.06 (30) (30) 0.0e 30k Hence m Thus e 30k 0.0, so 30k ln0.0 and hence k ln So b Exercise 46(ii), 3.34e.37.37e 0.009t in billions of people. ii. Observe from (i) that (t) 0.0e 0.009t. Taking t 0,,...,30 we get that the specific growth rates for 96, 970,...,99 are 0.0,0.00,0.09,0.08,0.08,0.07 and 0.06 respectivel. The approximation and Table.4 provides the specific growth rates 0.0,0.0,0.00,0.09,0.08,0.07 and 0.06 for the same sequence of ears. A look at Table.4 shows that Gompertz s model fits a bit better. iii. Return to the formula 3.34e.37.37e 0.009t. Taking t 3,, 8, and 30 gives 8

19 the projections (in billions) 6.,8.,.3,and 6.,respectivel,for the ears 000, 00,00,and 09. iv. Pushing t to infinit in the expression for gives 3.34e billion. J. Monod s Equation 49. i. Monod s equation specializes to 0.6s(t) 9+s(t). ii. With s(t) 0 +,we get t 0.6 [ 0 + ] t [0 + t ] 0t t +0 so that 3 t t t t t 0t +0, How do we solve this equation for? Start with a fact from Section 0.3. Because d ln dt (t) (t) we see that ln is an antiderivative of. If we can find an antiderivative of the right side of the equation,we can use the exponential function to solve for. As just asserted, ln +C t t + dt + t + dt. Because (see Section 0.) d dx tan x,we get that x + d t dt a b tan b a b + t b b a t +b b Taking b and a 3, we see that d 3 dt tan t 3,so t + 3 t t + dt 0.6 tan + C. Next,observe that [ d at a ] t b tan a ab dt b t + b a(t + b) ab t + b Taking b and a 0.06, 9 ab t + b. at t + b.

20 0.06 t t + dt 0.06t 0.06 tan + C. Combining the above conclusions, ln 0.06t 0.06 t tan +0.6 t tan + C. Therefore, ln 0.06t 0.4 tan ( t ) + C. B properties of the exponential function,we get e ln e 0.06t 0.4 tan ( t )+C e 0.06t 0.4 tan t If the formula s(t) 0 + is valid moments after the culture is first observed at time t t 0,then 0 (0) e 0 e C e C. Hence 0 e 0.06t 0.4 tan t. Let s have a look at the graph of the function. It is sketched below for 0 t e C das. Notice that it is essentiall a straight line. Because the term tan ( ) t bounded b π (see Figure 0.4 in Chapter 0.),it follows that for t large enough, is

21 e 0.06t will make the dominant contribution to the function. The graph of above for 0 t 60 reflects this. 0. i. Monod s equation sas that µmaxs(t). Letting a µ s / max +s(t) max and b s /max we can rewrite this as as(t). So we need to set b+s(t) f(t) as(t) b + s(t) and solve for s(t). Doing so,we get f(t)[b+s(t)] as(t),hence s(t)f(t) as(t) bf(t), and therefore s(t) bf(t) f(t) a bf(t) a f(t). So (t) f(t) can be achieved b taking s(t) bf(t) a f(t) ii. B equation (j), k µ and ( Me µt M M. So )+e 0 µt in Monod s equation. µ k. From the discussion in Section.B we know that Me µt µ µ ( ) µ µ ( ) M M 0 + e µt M e µt [ ] [ ] 0 e µt M e µt 0 e µt µ µ µ (M 0 )+ 0 e µt (M 0 )+ 0 e µt µ(m 0 ) (M 0 )+ 0 e µ µt + 0 M 0 e µt µe µt 0 M 0 + e. µt e µt So µe µt f(t) with f(t) 0. In reference to Monod s equation,note that s(t) +e M µt 0 e µt,µ max µ (which is the case b equation (j)),and s / max 0 M 0.