Section 7.4: Exponential Growth and Decay


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1 1 Sction 7.4: Exponntial Growth and Dcay Practic HW from Stwart Txtbook (not to hand in) p. 532 # 117 odd In th nxt two ction, w xamin how population growth can b modld uing diffrntial quation. W tart with th baic xponntial growth and dcay modl. Bfor howing how th modl ar t up, it i good to rcall om baic background ida from algbra and calculu. 1. A variabl y i proportional to a variabl x if y = k x, whr k i a contant. 2. Givn a function P(, whr P i a function of th tim t, th rat of chang of P with rpct to th tim t i givn by = P (. 3. A function P( i incraing ovr an intrval if = P ( >. A function P( i dcraing ovr an intrval if = P ( <. A function P( i nithr incraing or dcraing ovr an intrval if = P ( =. Th Exponntial Growth Modl Whn a population grow xponntially, it grow at a rat that i proportional to it iz at any tim t. Suppo th variabl P( (omtim w u jut u P) rprnt th population at any tim t. In addition, lt P b th initial population at tim t =, that i, P ( ) = P. Thn if th population grow xponntially, (Rat of chang of population at tim = k (Currnt population at tim In mathmatical trm, thi can b writtn a = kp. Solving for k giv k = 1 P Th valu k i known a th rlativ growth rat and i a contant.
2 2 Suppo w rturn to th quation = kp. W can olv thi quation uing paration of variabl. That i, = P 1 = P ln P = ln P P = = P( k k + C + C = C A (Sparat th variabl) (Intgrat both id) (Apply intgration formula) (Rai both id to xponntial function of (U invr proprty ln k (U abolut valu dfinition P = ± ba ) = k and law of xponnt b C x+ y = b b and rplac contant ± x y ) C with A.) Th quation P ( = A rprnt th gnral olution of th diffrntial quation. Uing th initial condition P ( ) = P, w can find th particular olution. P P A = P() = = A(1) = P A k() (Subtitut t = in th quation and quat to P (Not that k() (Solv for A) = = 1) ) Hnc, P ( = P i th particular olution. Summarizing, w hav th following: Exponntial Growth Modl Th initial valu problm for xponntial growth ha particular olution = kp, P() = P P ( = P whr P = initial population (population you that with) at tim t =, k = rlativ growth rat that i contant t = th tim th population grow. P( = what th population grow to aftr tim t.
3 3 Not[ 1. Whn modling a population with an xponntial growth modl, if th rlativ growth rat k i unknown, it hould b dtrmind. Thi i uually don uing th known population at two particular tim. 2. Exponntial growth modl ar good prdictor for mall population in larg population with abundant rourc, uually for rlativly hort tim priod. 3. Th graph of th xponntial quation ( = P ha th gnral form P P P ( P Exampl 1: Solv a crtain organim dvlop with a contant rlativ growth of.2554 pr mmbr pr day. Suppo th organim tart on day zro with 1 mmbr. Find th population iz aftr 7 day. Solution: t
4 4 Exampl 2: A population of a mall city had 3 popl in th yar 2 and ha grown at a rat proportional to it iz. In th yar 25 th population wa 37. a. Find an xprion for th numbr of popl in th city t yar aftr th yar 2. b. Etimat th population of th city in 26. In 21. c. Find th rat of growth of th population in 26. d. Auming th growth continu at th am rat, whn will th town hav 25 popl? Solution:
5 5
6 6 Exponntial Dcay Whn a population dcay xponntially, it dcra at a rat that i proportional to it iz at any tim t. Th modl for xponntial dcay i = kp, P( = P 1 Hr, k = i calld th rlativ dcay contant. Not that k > inc, bcau P 1 th population i dcraing, < and k = >. Uing paration of P ngativ ngativ variabl in a proc imilar to xponntial growth, it can b hown that th olution to th initial valu problm i P( P. Summarizing, w hav th following: = Exponntial Dcay Modl Th initial valu problm for xponntial dcay ha particular olution = kp, P() = P P( = P whr P = initial population (population you that with) at tim t =, k = rlativ dcay rat that i contant. Not that k >. t = th tim th population dcay. P( = th population that i lft aftr tim t. Not 1. Many tim th rat of dcay i xprd in trm of halflif, th tim it tak for half of any givn quantity to dcay o that only half of it original amount rmain. 2. Radioactiv lmnt typically dcay xponntially.
7 7 Exampl 3: Bimuth21 ha a halflif of 5. day. a. Suppo a ampl originally ha a ma of 8 mg. Find a formula for th ma rmaining aftr t day. b. Find th ma rmaining aftr 3 day. c. Whn i th ma rducd to 1 mg. d. Sktch th graph of th ma function. Solution: (Part a) Sinc thi i an xponntial dcay problm, w will u th formula P( = P. Sinc w tart with 8 mg, thn w know that P = 8. Thu th formula bcom P( = 8 To complt th quation that modl thi population, w nd to find th rlativ dcay rat k. W can u th half lif of th ubtanc to do thi. Th half lif of Bimuth21 i 5 day. Thi ay that aftr t = 5, th original population of 8 mg ha dcay to half of it original amount, 1 or (8) = 4 mg. Mathmatically, inc P ( rprnt that amount of population of th 2 ubtanc lft aftr tim t, thi ay that P ( 5) = 4. Uing th dcay quation, w hav or rarranging, w hav 4 = P(5) = k = 4 k(5) W mut olv thi quation for k. W procd with th following tp. ln 5k 5k 5k = 4 8 =.5 = ln(.5) 5k ln = ln(.5) 5k(1) = ln(.5) ln(.5) k = 5 k.1386 (Divid by id by 8) (Simplify) (Tak ln of both id) (U proprty ln b (Rcall ln = 1) = u ln b) (Divid both id of  5) u (U calculator and round to 4 dcimail plac) Subtituting k = and P 8 giv a formula for finding th rmaining ma. = P( = t (continud on nxt pag)
8 8 Part b.) Uing th formula.1386t P( = 8 found in part a, w that Ma rmaining.1386(3) = P(3) = 8 = 8 aftr t = 3 day 12.5 gram Part c.) In thi problm, w want th tim t it tak for th ma to hav rducd down to 1 mg. That i, w want t whn P ( = 1. W prform th following tp uing.1386t P( = 8 to olv for t. 8 1 = P( = t = t.1386t 1 = t 1 ln = ln t ln = ln t (1) = ln 8 ln( 1/ 8) t =.1386 t 48.2 (St P( = 1) (Rarrang th quation) (Divid both id by 8) (Tak ln of both id) (ln b u (ln = 1) = u lnb) (Divid both id by ) (U calculator to approximat) Thu, it tak approximatly t = day for th ubtanc to dcay to 1 mg. (Continud on nxt pag)
9 9 d. Th following Mapl command will gnrat th dird graph: > P := 8*xp(.1386*; P := 8 (.1386 t ) > plot(p, t =..6, color = blu, thickn = 2, viw = [ 1..6, 1..1], titl = "Graph of 8^( for t =..6 for Bimuth21");
10 1 Exampl 4: Radiocarbon Dating. Scintit can dtrmin th ag of ancint objct (foil, for xampl) uing radiocarbon dating. Th bombardmnt of th uppr atmophr by comic ray convrt nitrogn to a radioactiv iotop of carbon, C, with a half lif of about 573 yar. Vgtation aborb carbon dioxid through th atmophr and animal lif aimilat 14 C through food chain. Whn a plant or animal di, it top rplacing it carbon and th 14 amount of C bgin to dcra through radioactiv dcay. Thrfor, th lvl of radioactivity mut alo dcay xponntially. Suppo a foil found ha about 35 % a much 14 C Solution: radioactivity a normal animal do on Earth today. Etimat th ag of th foil. 14
11 11
12 12 Nwton Law of Cooling Nwton Law of Cooling tat that th rat of cooling of an objct i proportional to th diffrnc btwn th objct and it urrounding. Lt T ( b th tmpratur of an objct at tim t and T b th tmpratur of th urrounding (nvironmn. Not that T will w b aumd to b contant. Mathmatically, Nwton Law of Cooling can b xprd a th following diffrntial quation: dt = k ( T T Suppo w lt y = T T. Thn, taking th drivativ of both id with rpct to th dy dt dt dt dy tim t giv = = (rmmbr, T i contan. Subtituting = and y = T into th Nwton Law of cooling modl giv th quation T ) dy = k y. Thi i jut th baic xponntial growth modl. Th olution of thi diffrntial quation i y ( = y, whr y i th initial valu of y ( at tim t =, that i y ( ) = y. W lat nd to chang to olution back into an quation involving th tmpratur T. Rcall that y( = T. Uing thi quation, w that y = y( ) = T () T = T T (w u th variabl T to rprnt th initial tmpratur of th objct at tim t =, that i T ( ) = T. Subtituting y = T ( and y = T T into T = ( T T ) y ( = y giv Solving for T ( giv th olution to th Nwton Law of Cooling diffrntial quation: W ummariz th rult a follow: = T + ( T T )
13 13 Nwton Law of Cooling Th rat that th tmpratur T of an objct that i cooling i givn by th initial valu problm dt = k( T T ), T () = T. Th particular olution of thi initial valu problm dcribing th objct tmpratur aftr a particular tim t i givn by whr = T + ( T T ) T = th tmpratur of th urrounding nvironmnt. T = th initial tmpratur of th objct at tim t =. k = proportionality contant T ( = th tmpratur of th objct aftr tim t. W illutrat how thi law work in th nxt xampl.
14 14 Exampl 5: Suppo you cool a pot of oup in a 75 F room. Right whn you tak th oup off th tov, you maur it tmpratur to b 22 F. Suppo aftr 2 minut, th oup ha coold to 17 F. a. What will b th tmpratur of th oup in 3 minut. b. Suppo you can at th oup whn it i 13 F. How long will it tak to cool to thi tmpratur? dt Solution: (Part a) Th olution to initial valu problm = k( T T ), T () = T dcribing Nwton Law of Cooling i Sinc th tmpratur of th room i = T + ( T T ) 75, T = 75. Sinc th oup whn takn off th tov i 22, T = 22. Thu, th formula for dcribing th tmpratur aftr tim t i givn by or, whn implifid, = 75 + (22 75) = A with any xponntial modl, if th proportionality contant k, i not givn, w mut find it. To do thi, w u th fact that aftr t = 2 minut, th tmpratur i 17. Mathmatically, thi ay that T ( 2) = 17. W ubtitut thi fact into th abov quation and olv for k uing th following tp. 17 = T (2) = k 2k 2k = 17 = = 95 k(2) 2k 95 = 145 2k 95 ln = ln k ln = ln k(1) = ln 145 ln(95 /145) k = 2 k.211 (Rarrang and clan up th rulting quation) (Subtract 75 from both id) (Simplify) (Divid both id by 145) (Tak ln of both id) (U proprty ln b (ln = 1) (Divid both id by 2 to olv for (U calculator) u = u ln b) k) (Continud on nxt pag)
15 15 Subtituting k =. 211 into = giv = t Thu, w find th tmpratur aftr t = 3 ubtituting into thi quation. Hnc, Tmpratur of Soup = T (3) = aftr t = 3 minut = = 152 F.211(3).633 Part b) W want th tim t it tak for th oup to cool to 13, that i, th tim t whn.211t = 13. Taking th quation w found in part a = , w olv thi quation a follow: 13 = = t.211t.211t = 13 = = t (Rarrang th rulting quation) (Subtract 75 from both id) (Simplify).211t 55 = 145 (Divid both id by 145).211t 55 ln = ln 145 (Tak ln of both id) 55 u t ln = ln (U proprty ln b = u ln b) t (1) = ln 145 (ln = 1) ln(55 /145) t =.211 (Divid both id by to olv for t 46 (U calculator) Thu, it tak approximatly t = 46 minut for th oup to cool to 13 F.
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