Fall08SupplH - MAC 1105 Supplement H(5.6 The doubling time...

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Unformatted text preview: MAC 1105 - Supplement H (5.6) The doubling time, I of a population is: The function P{t) = Poe-k", P9 is the population at time t = 0 P is the population after time t is a model of man},r kinds of In 9 T2T- It is the exponential growth rate population growth. See Example 1 page 409. 1. In 2004. the world population was 6.4 billion. The exponential growth rate was 1.1 % per year and can be modeled with the exponential growth function: Pit] = 6.4 60m” Where it is the number of years since “2004 and P(t) is in billions. 3.) Estimate the population of the. world in 2006 and in 2016. b) When will the world population be 8 billion? 6) Find the doubling time. 2. The current natural population growth rate of the United States is 1.9% per year. The population in November of 2005 Was about 295.7 million . This growth can be modeled with the function: P(t) = 295.7 60.01:» Where 1? is the number of years since 2005 and P(t) is in millions. a) Estimate the population of the United States in 2006 and in 2011. b) W’hen will the United States population be 350 million? c) Find the doubling time. The doubling time, T of the amount is: P0 is the amount invested at time t z 0 The function P(t) = Poe“, is a. model of interest P is the amount after t years k- is the interest rate compounded continuously compounded continuously- See Example 2 pages 410-412. 3. Suppose that $10,000 is invested at an interest rate of 5.4 % compounded continuously. P(t) = 10, 000 (2-05“ a)_ What is the balance after 1 Year? 5 Years? 10 Years? b) “’hen will the account have $15,000? 6) Find the doubling time. (Revised 8/08) 41. Suppose that $60,000 is invested at an interest rate of 3.733 ‘70 compounded continuously. PM = 50, one 6'0375t a) What is the balance after 1 Year? 5 Years? 10 Years? b) When will the account have $100,000? C) Find the doubling time. P0 is the amount of a substance at time t = 0 The function PU) = Pas—M, k > 0 is a model of the decay P is the amount still radioactive after time t of a. radioactive substance. F; is a positive constant that is the decay rate. The half—life, T of a substance is the time that it takes for :19: of the substance to cease to be radioactive. See Example 5 page 415. 5. Poloninm has a Haifrlife T of 3 minutes. P{t) = P0 #3 3.) Find 1;, the decay rate. b) Write the equation Pit) with the constant k. 6. Lead has a Half—fife T of '22 years. P”) = P0 E—k22 a.) Find k, the decay rate. b) Write the equation PU) with the constant k. 7'. lodinelBl has a decay rate k- oi~ 9.6 % per day. Pm = P0 6496* 3,) Find T, the Half-life of Iodine-1371. {Revised 8/08) 1. U! a a) In 2006, t z 2 111 2016, t = 12 In 20.29 years in? , T — 11—11 — 63.01 In 2006.1: 1 In 201],t=6 In 8.8? years In 2 T = —-- = .48 .019 36 1 Year t m 1 5 Years t = 5 10 Years t = 10 In 7.5 years 1112 5 ”05—4 2 12.84 1 Year t = 1 5 Years 1 m 5 10 Years t = 10 In 13.62 ytéars In :2 T = = 18.48 .0375 k = .2310 MAC 1105 - Supplement H Answers (5.6) Pit) : 6.4 60-01191 = 6-54 Billion Pm = 6.4 eO'DUUQ) = 7.3 BEHion 01' about the middle of 2024 years. Pm = 295.? (90-9190) = 301.37 Million Pit) r. 29.5.7 60-019“) = 331.41 Minion Or towards the end of 2013 years . P(t) a 10, 000 60354“) = 3.10, 554.85 Pa) = 10, 000 £00545) = 313, 099.65 P(t) x: 10, 000 9-0-054‘10) = $17. 160.07 years. PU) = 60. 000 60-0315“) = $62, 292.72 P9) = 60. 000 9-9375“) : 972. 373.81 P(t) = 60, 009 6"“375‘40) = $87, 299.48 years. 23.1% per minute 3.15% per year (Revised '8 / 08) ...
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