and Decay Functions f (t) = C(1± r) t / K, for t 0, where

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1 MATH 116 Exponenial Growh and Decay Funcions Dr. Neal, Fall 2008 A funcion ha grows or decays exponenially has he form f () = C(1± r) / K, for 0, where C is he iniial amoun a ime 0: f (0) = C r is he rae of growh or decay in decimal, and 1 ± r is he base. K is he erm over which he growh or decay occurs. Example 1. Find he described exponenial growh or decay funcion. Graph he funcion and sae he domain and range. (a) Your savings S sars a $3000 and increases 15% every 4 years. (b) A baceria s mass M sars a 200 g and decreases 4% every 6 hours. (c) A child's weigh W sars a 50 lb. and gain 6% every 1/3 of a year for 7 years. Soluion. (a) Here C = 3000, r = 0.15, and K = 4. So we have S() = 3000(1.15) /4 for 0, in years. The range is [3000, ), or S Savings S in $ 3000 ime in yrs (b) Here we are decreasing exponenially, so we use he form C(1 r) / K, where C = 200, r = 0.04, and K = 6. Now we have M() = 200(0.96) /6 for 0, in hours. baceria mass M in g The range is (0, 200], or 0 < M 200. Noe: Because we are losing 4%, we also can say ha we are mainaining 96%. 200 ime in hrs

2 (c) A child's weigh W sars a 50 lb. and gain 6% every 1/3 of a year for 7 years. Here C = 50, r = 0.06, and K = 1/ 3. Noe now ha he exponen becomes / K = / (1/ 3) = 3. So we have W ( ) = 50(1.06) 3 for 0 7, in years. Then W (7) = 50(1.06) lbs The range is hen [50, 170], or 50 W 170. Alernae Form Weigh W in lbs ime in yrs Anoher form of he exponenial process is f () = Ca k where he base a = 1 ± r is deermined by a measuremen a anoher ime. Given he iniial saring amoun C and anoher measuremen ( D, K ) a a laer ime K, hen he funcion is given by f () = C ( D / C) / K for 0. Example 2. A small own s populaion is dropping exponenially. The iniial coun was 42,500. Bu wo years laer, i was down o 41,225. (i) Find he populaion decay funcion. (ii) Esimae he populaion coun for 20 years from he iniial coun. Soluion. The base is given by 41225/42500 = Tha is, he own is keeping only 97% of is populaion every 2 years (i.e., i is losing 3% every 2 years). The populaion decay funcion is hen P( ) = /2 for 0, where = 0 is he day of he iniial coun. Then P(20) = So 20 years afer he iniial coun (barring no change in circumsances), he own s populaion should be around 31,340. Solving an Exponenial Equaion Given f () = Ca /K, we may wan o solve he equaion f () = V. Here is he generic process o solve he equaion Ca /K = V : 1. Divide by C o isolae he base a / K = V / C 2. Take naural log of boh sides ln( a / K ) = ln(v / C) 3. Use fac ha ln a x = x ln a o drop exponen ln a = ln(v / C) K 4. Solve for = K ln V / C ( ) ln a

3 Example 3. A counry s populaion measured wih 250,000,000 people a he beginning of 2005 and is increasing 1.5% every half year. (a) Give he exponenial growh funcion. (b) Esimae he counry s populaion for he beginning of year (c) When will he populaion reach 400,000,000? Soluion. (a) Here C = , r = 0.015, and K = 1/ 2. Noe now ha he exponen becomes / K = / (1/ 2) = 2. So we have P( ) = ( 1.015) 2 for 0, where = 0 is he day of he iniial coun. (b) The beginning of year 2015 is 10 years from he iniial coun, and P(10) = ( 1.015) ,713,751.6, or abou 336,713,751 people. (c) Now we mus solve = ( 1.015) 2 = 400,000, Divide by = Take naural log ln ( ) = ln Drop exponen ln1.6 2 ln1.015 = ln Solve for = years. 2ln1.015 The beginning of years gives year monhs, or year monhs. Tha is, he counry s populaion should reach 400,000,000 by he middle of Sepember in Example 4. A he beginning of he semeser, your aenion span was 50 minues. Bu i drops 10% every week. In wha week will your aenion span be only 12 minues? Soluion. Your aenion span funcion is A( ) = , for 0, in weeks. Now we mus solve = 12. We hen have 0.90 = 0.24, and hen ln0.90 = ln0.24. So = ln0.24 / ln weeks from he sar of he semeser. So in he middle of he 14h week, your aenion span in class should be abou 12 minues. Half-Life Given an exponenial decay funcion f () = Ca /K, where 0 < a < 1, we can solve for he amoun of ime T H i akes for he process o cu in half. This ime is called he half-life. To find he half-life, we mus solve Ca /K = C / 2. The soluion is always given by T H = K ln( 0 / 5) ln a Once we have he half-life, we can re-wrie he funcion as f () = C 0.50 / T H

4 Example 5. The mass M of a subsance is decaying exponenially. The iniial mass was 6000 g. Five days laer, he amoun was 5340 g. (a) Give he exponenial decay funcion. (b) Describe he amoun of percenage decay in words. (c) Solve for he ime i akes for he mass o drop o 4% of is iniial amoun. (d) Give he half-life (o wo decimal places). Then wrie his exponenial decay funcion in erms of he half-life. Soluion. (a) The base is 5340/6000 = Thus, M() = /5, for 0, in days. (b) Saring from 6000 g, he mass is dropping 11% every 5 days. (c) Now we mus solve /5 = , or 0.89 /5 = By aking naural logs and dropping he exponen, we have 5ln(0.04) ln(0.89) = ln(0.04), and hen = 5 ln(0.89) days. (d) Now o find he half-life, he amoun mus drop o 50% of is iniial amoun. Thus we mus solve /5 = or 0.89 /5 = We hen obain a half-life of T H = = 5ln(0.50) days ln(0.89) We now can re-wrie he funcion as M() = /29.74 for 0, in days. Carbon-14 Daing The half-life of Carbon 14 is abou 5730 years. Every living organism has a cerain amoun of Carbon 14 in is sysem, and afer an organism dies, his Carbon 14 decays exponenially. By comparing he curren amoun of C 14 in a dead organism wih he amoun in a living organism, we can deermine how long ago he dead organism passed away. Example 6. In he year 2006, a human blood sample on a cloh conained 78.9% of is live-sae C14. Solve for he dae for which Carbon-14 daing places he blood sample. Soluion. Le f () = C (0.50) /5730, wih in years, where C is he live-sae amoun of C 14. We now solve for he ime i akes o drop o C (i.e., 78.9% of live-sae). C (0.50) /5730 = C (0.50) /5730 = ln(0.50) = ln(0.789) 5730 = 5730ln(0.789) ln(0.50) years ago. So he dae of deah is A.D.

5 Exercises 1. Find he exponenial growh or decay funcion: (a) Sar wih amoun 40,000 and decrease 12% every 6 years. (b) Sar wih amoun 2000 and increase 7.5% every half of a year. (c) Sar wih amoun 50. In 10 hours, he amoun is 120. (d) Sar wih amoun In 4 days, he amoun is The U.S. populaion is growing exponenially. On July 1, 1990 he pop. was abou 250 million. On July 1, 1995 he pop. was abou 263 million. Assuming his rae says he same, (a) Find a funcion for he exponenial growh (saring on July 1, 1990). (b) Explain he growh rae in words. (c) Esimae he pop. on July 1, (d) How long unil he U.S. populaion reaches 500 million? 3. You deposi $5000. Every quarer (1/4 h of a year) you gain 2%. (a) Find a funcion for he exponenial growh. (b) Esimae he amoun hree years. (c) How long unil he accoun reaches $12,000?

6 Soluions 1. (a) Decrease 12% mainain 88% f () = (0.88) /6, 0, in years. (b) f () = 2000 (1.075) /(1/2) = 2000 (1.075) 2, 0, in years. (c) f () = 50 (120 / 50) /10 = 50 (2.4) /10, 0, in hours. (d) f () = 3000 (2400/ 3000) /4 = 3000 (0.8) /4, 0, in days The base is / = So, P( ) = (1.052) /5, 0, in years, where = 0 represens July 1, (b) Saring wih 250,000,000 on July 1, 1990, he populaion is growing 5.2% every 5 years. (c) July 1, 2050 is 60 years from sar, and P(60) = (1.052) ,334,310. (d) Solve (1.052) /5 = (1.052) /5 = 2 ln(1.052) = ln2 5 5ln2 = years from July 1, This dae is hen July 1, ln(1.052) monhs, or July 1, monhs, which is he middle of Nov (a) f () = 5000 (1.02) 4, 0, in years. (b) f (3) = 5000 (1.02) 12 $ (c) 5000 (1.02) 4 = (1.02) 4 = ln(1.02) = ln(2.4) = years. ln(2.4) 4 ln(1.02)