Exponential Growth and Decay; Modeling Data


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1 Exponntial Growth and Dcay; Modling Data In this sction, w will study som of th applications of xponntial and logarithmic functions. Logarithms wr invntd by John Napir. Originally, thy wr usd to liminat tdious calculations involvd in multiplying, dividing, and taking powrs and roots of th larg numbrs that occur in diffrnt scincs. Computrs and calculators hav sinc liminatd th nd of logarithms for ths calculations. Howvr, thir nd has not bn rmovd compltly. Logarithms aris in problms of xponntial growth and dcay bcaus thy ar invrss of xponntial functions. Bcaus of th Laws of Logarithms, thy also turn out to b usful in th masurmnt of th loudnss of sounds, th intnsity of arthquaks, and othr procsss that occur in natur. Prviously, w studid th formula for xponntial growth, which modls th growth of animal or bactria population. If n is th initial siz of a population xprincing xponntial growth, thn th population n(t) at tim t is modld by th function nt () n rt whr r is th rlativ rat of growth xprssd as a fraction of th population. Now, w hav th powrful logarithm, which will allow us to answr qustions about th tim at which a population rachs a crtain siz. Exampl 1: Frog population projctions Th frog population in a small pond grows xponntially. Th currnt population is 85 frogs, and th rlativ growth rat is 18% pr yar. (a) Find a function that modls th numbr of frogs aftr t yars. (b) Find th projctd population aftr 3 yars. (c) Find th numbr of yars rquird for th frog population to rach 6. Solution (a): To find th function that modls population growth, w nd to find th population n(t). To do this, w us th formula for population growth with n 85 and r.18. Thn nt () 85 By Jnnifr Guajardo
2 Exampl 1 (Continud): Solution (b): W us th population growth function found in (a) with t 3. n(3) 85.18(3) n(3) Us a calculator Thus, th numbr of frogs aftr 3 yars is approximatly 146. Solution (c): Using th function w found in part (a) with n(t) 6 and solving th rsulting xponntial quation for t, w gt ln(7.59) ln(7.59) t.18 t 1.86 ln( ) ln(7.59) Divid by 85 Tak ln of ach sid Proprty of ln Divid by.18 Us a calculator Thus, th frog population will rach 6 in approximatly 1.9 yars. Exampl 2: Find th numbr of bactria in a cultur A cultur contains 1, bactria initially. Aftr an hour, th bactria count is 25,. (a) (b) Find th doubling priod. Find th numbr of bactria aftr 3 hours. Solution (a): W nd to find th function that modls th population growth, n(t). In ordr to find this, w must first find th rat r. To do this, w us th formula for population growth with n 1,, t 1, and n(t) 25,, and thn solv for r. r (1) 1, 25, r 2.5 r ln( ) ln(2.5) r ln(2.5) r Divid by 1, Tak ln of ach sid Proprty of ln Us a calculator By Jnnifr Guajardo
3 Exampl 2 (Continud): Now that w know r.91629, w can writ th function for th population growth: nt ( ) 1, Rcall that th original qustion is to find th doubling priod, so w ar not don yt. W nd to find th tim, t, whn th population n(t) 2,. W us th population growth function found abov and solv th rsulting xponntial quation for t. 1, 2, 2 ln( ) t t.756 Divid by 1, Tak ln of ach sid Proprty of ln Divid by Us a calculator Thus, th bactria count will doubl in about.75 hours. Solution (b): Using th population growth function found in part (a), with rat r and tim t 3, w find n(3) 1,.91629(3) 156, Us a calculator Radioactiv Dcay: So, th numbr of bactria aftr 3 hours is about 156,25. In radioactiv substancs th mass of th substanc dcrass, or dcays, by spontanously mitting radiation. Th rat of dcay is dirctly proportional to th mass of th substanc. Th amount of mass m(t) rmaining at any givn tim t can b shown to b modld by th function mt () m rt By Jnnifr Guajardo
4 whr m is th initial mass and r is th rat of dcay. In gnral, physicists xprss th rat of dcay in trms of halflif, th tim rquird for half th mass to dcay. Somtims, w ar givn th halflif valu and nd to find th rat of dcay. To obtain this rat, follow th nxt fw stps. Lt h rprsnt th halflif and assum that our initial mass is 1 unit. This forcs m(t) to b ½ unit whn t h. Now, substituting all of this information into our modl, w gt ln rh 2 rh r h r h 1 ln(2 1 ) Tak ln of ach sid Solv for r 1 ln(2 )  by law 3 This quation for r will allow us to find th rat of dcay whnvr w ar givn th halflif h. If m is th initial mass of a radioactiv substanc with halflif h, thn th mass rmaining at tim t is modld by th function: mt () m rt whr r. h Exampl 3: Radioactiv Dcay Th halflif of csium137 is 3 yars. Suppos w hav a 1 g sampl. (a) (b) (c) (d) Find a function that modls th mass rmaining aftr t yars. How much of th sampl will rmain aftr 8 yars? Aftr how long will only 2 g of th sampl rmain? Draw a graph of th sampl mass as a function of tim. Solution (a): Using th modl for radioactiv dcay with m 1 and r.2315, w hav: 3 ( ) mt () t By Jnnifr Guajardo
5 Exampl 3 (Continud): Solution (b): W us th function w found in part (a) with t 8. m.2315(8) (8) Thus, approximatly 1.6 g of csium137 rmains aftr 8 yars. Solution (c): W us th function w found in part (a) with m(t) 2 and solv th rsulting xponntial quation for t..2315t t t ln( ) ln.2315t ln 5 ln 5 t.2315 t 69.7 Divid by 1 Tak ln of ach Proprty of ln sid Divid by Us a calculator Th tim rquird for th sampl to dcay to 2 g is about 7 yars. Solution (d): A graph of th function mt () t is shown blow. By Jnnifr Guajardo
6 Nwton s Law of Cooling: Nwton s Law of Cooling stats that th rat of cooling of an objct is proportional to th tmpratur diffrnc btwn th objct and its surroundings, providd that th tmpratur diffrnc is not too larg. Using calculus, th following modl can b dducd from this law. If D is th initial tmpratur diffrnc btwn an objct and its surroundings, and if its surroundings hav tmpratur T S, thn th tmpratur of th objct at tim t is modld by th function Tt () T+ D S kt whr k is a positiv constant that dpnds on th typ of objct. Exampl 4: Nwton s Law of Cooling is usd in homicid invstigations to dtrmin th tim of dath. Th normal body tmpratur is 98.6 F. Immdiatly following dath, th body bgins to cool. It has bn dtrmind xprimntally that th constant in Nwton s Law of Cooling is approximatly k.1947, assuming tim is masurd in hours. Suppos that th tmpratur of th surroundings is 6 F. (a) (b) (c) Find a function T(t) that modls th tmpratur t hours aftr dath. If th tmpratur of th body is now 72 F, how long ago was th tim of dath? Find th tmpratur of th body aftr 9 hours. Solution (a): Th tmpratur of th surroundings is T S 6 F, and th initial tmpratur diffrnc is D F So, by Nwton s Law of Cooling and th givn constant valu k.1947, th tmpratur aftr t hours is modld by th function Tt ( ) t Solution (b): W us th function w found in part (a) with T(t) 72 and solv th rsulting xponntial quation for t. By Jnnifr Guajardo
7 Exampl 4 (Continud): t t t t ln(.3188) ln(.3188) t.1947 t 6.7 Subtract 6 Divid by 38.6 Tak ln of ach sid Divid by Us a calculator Solution (c): W us th function w found in part (a) with t 9. T.1947(9) (9) F Thus, th tmpratur of th body aftr 9 hours will b approximatly 66.7 F. By Jnnifr Guajardo
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