4.8 Exponential Growth and Decay; Newton s Law; Logistic Growth and Decay

Transcription

1 324 CHAPTER 4 Exponenial and Logarihmic Funcions 4.8 Exponenial Growh and Decay; Newon s Law; Logisic Growh and Decay OBJECTIVES 1 Find Equaions of Populaions Tha Obey he Law of Uninhibied Growh 2 Find Equaions of Populaions Tha Obey he Law of Decay 3 Use Newon s Law of Cooling 4 Use Logisic Models 1 Find Equaions of Populaions Tha Obey he Law of Uninhibied Growh Many naural phenomena have been found o follow he law ha an amoun A varies wih ime according o

2 SECTION 4.8 Exponenial Growh and Decay; Newon s Law; Logisic Growh and Decay 32 A = A e k (1) Here A is he original amoun 1 = 2 and k Z is a consan. If k 7, hen equaion (1) saes ha he amoun A is increasing over ime; if k 6, he amoun A is decreasing over ime. In eiher case, when an amoun A varies over ime according o equaion (1), i is said o follow he exponenial law or he law of uninhibied growh 1k 7 2 or decay 1k 6 2. See Figure 2. Figure 2 A A A A (a) A( ) A e k, k (b) A( ) A e k, k For example, we saw in Secion 4.7 ha coninuously compounded ineres follows he law of uninhibied growh. In his secion we shall look a hree addiional phenomena ha follow he exponenial law. Cell division is he growh process of many living organisms, such as amoebas, plans, and human skin cells. Based on an ideal siuaion in which no cells die and no by-producs are produced, he number of cells presen a a given ime follows he law of uninhibied growh. Acually, however, afer enough ime has passed, growh a an exponenial rae will cease due o he influence of facors such as lack of living space and dwindling food supply. The law of uninhibied growh accuraely reflecs only he early sages of he cell division process. The cell division process begins wih a culure conaining N cells. Each cell in he culure grows for a cerain period of ime and hen divides ino wo idenical cells. We assume ha he ime needed for each cell o divide in wo is consan and does no change as he number of cells increases. These new cells hen grow, and evenually each divides in wo, and so on. Uninhibied Growh of Cells A model ha gives he number N of cells in a culure afer a ime has passed (in he early sages of growh) is N12 = N e k, k 7 (2) where N is he iniial number of cells and k is a posiive consan ha represens he growh rae of he cells. In using formula (2) o model he growh of cells, we are using a funcion ha yields posiive real numbers, even hough we are couning he number of cells, which mus be an ineger. This is a common pracice in many applicaions.

3 326 CHAPTER 4 Exponenial and Logarihmic Funcions EXAMPLE 1 Bacerial Growh A colony of baceria grows according o he law of uninhibied growh according o he funcion N12 = 1e.4, where N is measured in grams and is measured in days. (a) Deermine he iniial amoun of baceria. (b) Wha is he growh rae of he baceria? (c) Graph he funcion using a graphing uiliy. (d) Wha is he populaion afer days? (e) How long will i ake for he populaion o reach 14 grams? (f) Wha is he doubling ime for he populaion? Soluion (a) The iniial amoun of baceria, N, is obained when =, so N = N12 = 1e.412 = 1 grams. Figure (b) Compare N12 = 1e.4 o N12 = 1e k. The value of k,.4, indicaes a growh rae of 4.%. (c) Figure 3 shows he graph of N12 = 1e.4. (d) The populaion afer days is N12 = 1e.412 = 12.2 grams. (e) To find how long i akes for he populaion o reach 14 grams, we solve he equaion N12 = 14. 1e.4 = 14 e.4 = = ln 1.4 ln 1.4 =.4 L 7. days Divide boh sides of he equaion by 1. Rewrie as a logarihm. Divide boh sides of he equaion by.4. (f) The populaion doubles when N12 = 2 grams, so we find he doubling ime by solving he equaion 2 = 1e.4 for. 2 = 1e.4 2 = e.4 Divide boh sides of he equaion by 1. ln 2 =.4 Rewrie as a logarihm. = ln 2 Divide boh sides of he equaion by.4..4 L 1.4 days The populaion doubles approximaely every 1.4 days. NOW WORK PROBLEM 1. EXAMPLE 2 Bacerial Growh A colony of baceria increases according o he law of uninhibied growh. (a) If he number of baceria doubles in 3 hours, find he funcion ha gives he number of cells in he culure. (b) How long will i ake for he size of he colony o riple? (c) How long will i ake for he populaion o double a second ime (ha is, increase four imes)?

4 SECTION 4.8 Exponenial Growh and Decay; Newon s Law; Logisic Growh and Decay 327 Soluion (a) Using formula (2), he number N of cells a a ime is k N12 = N e where N is he iniial number of baceria presen and k is a posiive number. We firs seek he number k. The number of cells doubles in 3 hours, so we have Bu N132 = N e k132, so N132 = 2N N e k132 = 2N e 3k = 2 3k = ln 2 Divide boh sides by N. Wrie he exponenial equaion as a logarihm. k = 1 3 ln 2 Formula (2) for his growh process is herefore (b) The ime needed for he size of he colony o riple requires ha N = 3N. We subsiue 3N for N o ge a 1 ln 2b = ln 3 3 a N12 = N e 1 ln 2b 3 a 3N = N e 1 ln 2b 3 a 3 = e 1 ln 2b 3 = 3 ln 3 ln 2 L 4.7 hours I will ake abou 4.7 hours or 4 hours, 4 minues for he size of he colony o riple. (c) If a populaion doubles in 3 hours, i will double a second ime in 3 more hours, for a oal ime of 6 hours. 2 Find Equaions of Populaions Tha Obey he Law of Decay Radioacive maerials follow he law of uninhibied decay. Uninhibied Radioacive Decay The amoun A of a radioacive maerial presen a ime is given by A12 = A e k, k 6 (3) A where is he original amoun of radioacive maerial and k is a negaive number ha represens he rae of decay. All radioacive subsances have a specific half-life, which is he ime required for half of he radioacive subsance o decay. In carbon daing, we use he fac ha all living organisms conain wo kinds of carbon, carbon 12 (a sable carbon) and carbon 14 (a radioacive carbon wih a half-life of 6 years). While an organism is living, he

5 328 CHAPTER 4 Exponenial and Logarihmic Funcions raio of carbon 12 o carbon 14 is consan. Bu when an organism dies, he original amoun of carbon 12 presen remains unchanged, whereas he amoun of carbon 14 begins o decrease. This change in he amoun of carbon 14 presen relaive o he amoun of carbon 12 presen makes i possible o calculae when an organism died. EXAMPLE 3 Soluion Esimaing he Age of Ancien Tools Traces of burned wood along wih ancien sone ools in an archeological dig in Chile were found o conain approximaely 1.67% of he original amoun of carbon 14. (a) If he half-life of carbon 14 is 6 years, approximaely when was he ree cu and burned? (b) Using a graphing uiliy, graph he relaion beween he percenage of carbon 14 remaining and ime. (c) Deermine he ime ha elapses unil half of he carbon 14 remains. This answer should equal he half-life of carbon 14. (d) Use a graphing uiliy o verify he answer found in par (a). (a) Using formula (3), he amoun A of carbon 14 presen a ime is A where is he original amoun of carbon 14 presen and k is a negaive number. We firs seek he number k. To find i, we use he fac ha afer 6 years half of he original amoun of carbon 14 remains, so Then, 1 2 A = A e k = e6k k A12 = A e A162 = 1 2 A. Divide boh sides of he equaion by A. 6k = Rewrie as a logarihm. Formula (3) herefore becomes If he amoun A of carbon 14 now presen is 1.67% of he original amoun, i follows ha.167a = A e 6 k = 1 6 L A12 = A e = e 6 Divide boh sides of he equaion by A. = ln = 6 ln.167 L 33,62 years Rewrie as a logarihm.

6 SECTION 4.8 Exponenial Growh and Decay; Newon s Law; Logisic Growh and Decay 329 Figure 4 1 The ree was cu and burned abou 33,62 years ago. Some archeologiss use his conclusion o argue ha humans lived in he Americas 33, years ago, much earlier han is generally acceped. (b) Figure 4 shows he graph of y = e where y is he fracion of carbon 14 presen and x is he ime. 6 x, (c) By graphing Y 1 =. and Y 2 = e where x is ime, and using INTERSECT, we find ha i akes 6 years unil half he carbon 14 remains. The half-life of carbon 14 is 6 years. 6 x, 4, 6 (d) By graphing Y and Y 2 = e x 1 =.167, where x is ime, and using INTERSECT, we find ha i akes 33,62 years unil 1.67% of he carbon 14 remains. NOW WORK PROBLEM 3. 3 Use Newon s Law of Cooling Newon s Law of Cooling * saes ha he emperaure of a heaed objec decreases exponenially over ime oward he emperaure of he surrounding medium. Newon s Law of Cooling The emperaure u of a heaed objec a a given ime can be modeled by he following funcion: u12 = T + 1u - T2e k, k 6 (4) is he ini- where T is he consan emperaure of he surrounding medium, ial emperaure of he heaed objec, and k is a negaive consan. u EXAMPLE 4 Using Newon s Law of Cooling An objec is heaed o 1 C (degrees Celsius) and is hen allowed o cool in a room whose air emperaure is 3 C. (a) If he emperaure of he objec is 8 C afer minues, when will is emperaure be C? (b) Using a graphing uiliy, graph he relaion found beween he emperaure and ime. (c) Using a graphing uiliy, verify ha afer 18.6 minues he emperaure is C. (d) Using a graphing uiliy, deermine he elapsed ime before he objec is 3 C. (e) Wha do you noice abou he emperaure as ime passes? Soluion (a) Using formula (4) wih T = 3 and u = 1, he emperaure (in degrees Celsius) of he objec a ime (in minues) is u12 = e k = 3 + 7e k * Named afer Sir Isaac Newon ( ), one of he cofounders of calculus.

7 33 CHAPTER 4 Exponenial and Logarihmic Funcions where k is a negaive consan. To find k, we use he fac ha u = 8 when =. Then 8 = 3 + 7e k12 = 7e k e k = 7 k = Formula (4) herefore becomes We wan o find when u = C, so e k = 1 L = 7e = 2 7 u12 = 3 + 7e = 3 + 7e = ln 2 7 = 2 ln 7 L 18.6 minues Figure 1 The emperaure of he objec will be C afer abou 18.6 minues or 18 minues, 37 seconds. (b) Figure shows he graph of y = 3 + 7e x is he ime. where y is he emperaure and (c) By graphing Y1 = and Y 2 = 3 + 7e where x is ime, and using INTER- SECT, we find ha i akes x = 18.6 minues (18 minues, 37 seconds) for he emperaure o cool o C. (d) By graphing Y 1 = 3 and Y 2 = 3 + 7e where x is ime, and using INTER- SECT, we find ha i akes x = minues (39 minues, 13 seconds) for he emperaure o cool o 3 C. x (e) As x increases, he value of e approaches zero, so he value of y, he emperaure of he objec, approaches 3 C, he air emperaure of he room. x, x, x, NOW WORK PROBLEM Use Logisic Models The exponenial growh model A12 = A e k, k 7, assumes uninhibied growh, meaning ha he value of he funcion grows wihou limi. Recall ha we saed

8 SECTION 4.8 Exponenial Growh and Decay; Newon s Law; Logisic Growh and Decay 331 ha cell division could be modeled using his funcion, assuming ha no cells die and no by-producs are produced. However, cell division would evenually be limied by facors such as living space and food supply. The logisic model can describe siuaions where he growh or decay of he dependen variable is limied. The logisic model is given nex. Logisic Model In a logisic growh model, he populaion P afer ime obeys he equaion P12 = c 1 + ae -b () where a, b, and c are consans wih c 7. The model is a growh model if b 7 ; he model is a decay model if b 6. Figure 6 The number c is called he carrying capaciy (for growh models) because he value P12 approaches c as approaches infiniy; ha is, lim P12 = c. The number : q ƒbƒ is he growh rae for b 7 and he decay rae for b 6. Figure 6(a) shows he graph of a ypical logisic growh funcion, and Figure 6(b) shows he graph of a ypical logisic decay funcion. P( ) y c P( ) (, P()) y c 1 c 2 Inflecion poin 1 c 2 Inflecion poin (, P()) (a) (b) Based on he figures, we have he following properies of logisic growh funcions. Properies of he Logisic Growh Funcion, Equaion () 1. The domain is he se of all real numbers. The range is he inerval 1, c2, where c is he carrying capaciy. 2. There are no x-inerceps; he y-inercep is P There are wo horizonal asympoes: y = and y = c. 4. P12 is an increasing funcion if b 7 and a decreasing funcion if b There is an inflecion poin where P12 equals of he carrying capaciy. 2 The inflecion poin is he poin on he graph where he graph changes from being curved upward o curved downward for growh funcions and he poin where he graph changes from being curved downward o curved upward for decay funcions. 6. The graph is smooh and coninuous, wih no corners or gaps.

9 332 CHAPTER 4 Exponenial and Logarihmic Funcions EXAMPLE Frui Fly Populaion Frui flies are placed in a half-pin milk bole wih a banana (for food) and yeas plans (for food and o provide a simulus o lay eggs). Suppose ha he frui fly populaion afer days is given by P12 = (a) Sae he carrying capaciy and he growh rae. (b) Deermine he iniial populaion. (c) Use a graphing uiliy o graph P12. (d) Wha is he populaion afer days? (e) How long does i ake for he populaion o reach 18? (f) How long does i ake for he populaion o reach one-half of he carrying capaciy? e -.37 Figure 7 2 Soluion (a) As : q, e -.37 : and P12 : >1. The carrying capaciy of he half-pin bole is frui flies. The growh rae is ƒbƒ = ƒ.37ƒ = 37%. (b) To find he iniial number of frui flies in he half-pin bole, we evaluae P12. So iniially here were 4 frui flies in he half-pin bole. (c) See Figure 7 for he graph of P12. (d) To find he number of frui flies in he half-pin bole afer days, we evaluae P12. P12 = P12 = = = e L 23 frui flies e 2 Afer days, here are approximaely 23 frui flies in he bole. (e) To deermine when he populaion of frui flies will be 18, we solve he equaion e -.37 = 18 = e = e = 6.e = e -.37 ln1.492 = -.37 L 14.4 days Divide boh sides by 18. Subrac 1 from boh sides. Divide boh sides by 6.. Rewrie as a logarihmic expression. Divide boh sides by I will ake approximaely 14.4 days (14 days, 9 hours) for he populaion o reach 18 frui flies.

10 SECTION 4.8 Exponenial Growh and Decay; Newon s Law; Logisic Growh and Decay 333 We could also solve his problem by graphing Y 1 = Y 2 = 18 and using INTERSECT. See Figure e -.37 (f) One-half of he carrying capaciy is 11 frui flies. We solve P12 = 11 by graphing Y 1 = and Y 2 = 11 and using INTERSECT. See e -.37 Figure 9. The populaion will reach one-half of he carrying capaciy in abou 1.9 days (1 days, 22 hours). and Figure 8 Figure 9 2 Y e.37 Y Y e.37 Y Exploraion On he same viewing recangle, graph Y 1 = and Y 2 = e e -.8. Wha effec does he growh rae ƒbƒ have on he logisic growh funcion? Look back a Figure 9. Noice he poin where he graph reaches 11 frui flies (one-half of he carrying capaciy): he graph changes from being curved upward o being curved downward. Using he language of calculus, we say he graph changes from increasing a an increasing rae o increasing a a decreasing rae. For any logisic growh funcion, when he populaion reaches one-half he carrying capaciy, he populaion growh sars o slow down. NOW WORK PROBLEM 21. EXAMPLE 6 Wood Producs The EFISCEN wood produc model classifies wood producs according o heir life-span. There are four classificaions; shor (1 year), medium shor (4 years), medium long (16 years), and long ( years). Based on daa obained from he European Fores Insiue, he percenage of remaining wood producs afer years for wood producs wih long life-spans (such as hose used in he building indusry) is given by P12 = e.81 (a) Wha is he decay rae? (b) Use a graphing uiliy o graph P12. (c) Wha is he percenage of remaining wood producs afer 1 years? (d) How long does i ake for he percenage of remaining wood producs o reach percen? (e) Explain why he numeraor given in he model is reasonable.

11 334 CHAPTER 4 Exponenial and Logarihmic Funcions Figure 6 11 Soluion (a) The decay rae is ƒbƒ = ƒ -.81ƒ =.81%. (b) The graph of P12 is given in Figure 6. (c) Evaluae P112. P112 = L e 1 So 9% of wood producs remain afer 1 years. (d) Solve he equaion P12 = e.81 = = e = e =.316e = e.81 ln =.81 L 9.6 years Divide boh sides by. Subrac 1 from boh sides. Divide boh sides by.316. Rewrie as a logarihmic expression. Divide boh sides by.81. I will ake approximaely 9.6 years for he percenage of wood producs remaining o reach %. (e) The numeraor of is reasonable because he maximum percenage of wood producs remaining ha is possible is 1%. NOW WORK PROBLEM 27.