Section 6.8. Solutions

Section 6.8: Solutions 1) The number of text messages (in billions) sent in the United States in a given year obeys the model N(t) = 12e.56t (where t = 0 represents the number of text messages sent in the year 2012.) 1a) Determine the number of text messages sent in the year 2012. Just plug in 0 for t and use your calculator. N(0) = 12e.56*0 N(0) = 12 Answer: 12 billion text messages will be sent in the year 2012 1b) What is the growth rate of text messages in the United States? The growth rate in decimal form is the .56 next to the t in the exponent. I will move the decimal two places to the right to create a percent. Answer: 56% 1c) How many text messages will be sent in the year 2016? t = 1 stands for 2013, t = 2 for 2014, I will put 4 into the equation for t to answer this question. N(4) = 12e.56*4 N(4) = 112.72 Answer: 112.72 billion text messages will be sent in the year 2016. 1d) How long will it take for the number of text messages to triple? I will substitute 3*12 = 36 in for N(t) and solve for t. 36 = 12e.56t (divide each side by 12) 3 = e.56t (take ln of each side) Ln 3 = ln e.56t (use power to product rule) Ln 3 = .56tln e (ln e = 1) Ln(3) = .56t (divide by .56, then use calculator) Answer: it will take 1.96 years for the number of text messages to triple. 3) The size P of a certain insect population in time t (in days) obeys the model P(t) = 200e.04t 3a) Determine the number of insects at t = 0 days Just substitute 0 for t in the formula P(0) = 200e.04*0 P(0) = 200 Answer: the initial population is 200 insects 3b) What is the growth rate of the insect population? The growth rate as a decimal is .04, move the decimal over two places to make this a percent. Answer: The growth rate is 4% 3c) What will the population be after 7 days? Substitute 7 for t, P(7) = 200e.04*7 P(7) = 265 ( I rounded to the nearest insect as there can’t be fractional insects) Answer: there will be 265 insects after 7 days. 3d) When will the population reach 3,000 insects 3000 = 200e.04t (now solve for t) (divide both sides by 300) 15 = e.04t (take ln of each side) Ln(15) = ln(e.04t) (use power to product rule) Ln(15) = .04tln e ( ln e = 1) Ln(15) = .04t (divide by .04 then use calculator) Answer: 67.70 days 3e) How long will it take for the population to double? Replace the P(t) with 400 and solve for t 400 = 200e.04t (divide by 200) 2 = e.04t (take ln of each side) Ln 2 = ln e.04t (use power to product rule) Ln 2 = .04tln e (ln e = 1) Ln 2 = .04t (divide by .04 then use calculator) Answer: 17.33 years 5) Krypton-85 is a radioactive material that decays according to the function P(t) = P0e-.063t where P0 is the initial amount present and P(t) is the amount present after t years. Assume a scientist has a sample of 1000 grams of Krypton-85. 5a) What is the decay rate of Krypton-85 ? The decay rate is the -063 or 6.3%, Answer: decay rate -6.3% per year 5b) How much Krypton-85 will be left after 30 years? Plug in 30 for t and 1000 for P0 (the initial amount) P(30) = 1000e-.063*30 P(30) = 151.07 Answer: 151.07 years 5c) When will 200 grams of Krypton-85 be left? Replace the left side of the equation with 200 and solve for t. 200 = 1000e-.063t (divide by 1000) .2 = e-.063t (take ln of each side) Ln .2 = ln e-.063t (use power to product rule) Ln .2 = -.063t ln e (ln e = 1) Ln .2 = -.063t Answer: 25.55 years 5d) What is the half-life of Krypton-85 ? Replace the left side of the equation with 500 and solve for t. 500 = 1000e-.063t (divide by 1000) .5= e-.063t (take ln of each side) Ln .5 = ln e-.063t (use power to product rule) Ln .5 = -.063t ln e (ln e = 1) Ln .5 = -.063t Answer: 11 years #7-10: Use the formula P(t) = P0ekt , where P0 is the initial population at t=0, and k is the rate of growth. (Round k to 3 decimals.) 7) The rodent population in a given city rose from 20,000 to 30,000 in 1 year. How long will it take the population to reach 100,000? (round to 2 decimals) First find k by replacing P(t) with 30,000, P0 with 20,000 and t = 1 (all values are from first sentence) 30000 = 20000ek*1 (divide by 20000) 1.5 = ek (take ln of both sides) Ln 1.5 = ln ek (use power to product rule) Ln 1.5 = k ln e Ln 1.5 = k k = .405 Now I can answer the question, replace the P(t) with 100,000 and k with .405 and solve for t 100,000 = 20,000e.405t (divide by 20,000) 5 = e.405t (take ln of each side) Ln 5 = ln e.405t Ln 5 = .405t ln e Ln 5 = .405t Answer: 3.97 years 9) The population of a city is expected to triple in 20 years. The city currently has 10,000 residents. How long will it take to get to 50,000 residents? (round to 2 decimals) First find k by replacing P(t) with 30,000, P0 with 10,000 and t = 20 (all values are from first two sentences) 30000 = 10000ek*20 (divide by 10000) 3= e20k (take ln of both sides) Ln 3 = ln e20k (use power to product rule) Ln 3 = 20k ln e Ln 3 = 20k k = .055 Now I can answer the question, replace the P(t) with 50,000 and k with .055 and solve for t 50,000 = 10,000e..055t (divide by 10,000) 5 = e.055t (take ln of each side) Ln 5 = ln e.055t Ln 5 = .055t ln e Ln 5 = .055t Answer: 29.26 years #11-14- Use the formula: , where P(t) = P0ekt is the amount present at time t = 0, 11) The half life of a certain element is 10 years. A 20 ounce sample of the element is obtained. How long will it take for there to be 12 ounces of the sample left? (round to 2 decimals) I will put t = 10 P0= 20 and P(t) = 10 ( the P(t) is half of the 20 ounce sample) 10= 20ek*10 (divide by 20) .5= ek*10 (take ln of each side) Ln .5= ln e10k (use power to product rule) Ln .5 = 10k ln e Ln .5 = 10k k = -.069 now replace P(t) with 12, and k with -.069 and solve for t. 12 = 20e-069t (divide by 20) .6 = e-.069t (take ln of each side) Ln .6 = ln e-.069t Ln .6 = -.069t lne Ln .6 = -.069t Answer: 7.40 years 13) The half life of a certain element is 5 years. A 60 ounce sample of the element is obtained. How long will it take until 15 ounces of the sample are left? (round to two decimals) I will put t = 5 P0= 60 and P(t) = 30 ( the P(t) is half of the 60 ounce sample) 30= 60ek*5 (divide by 60) .5= ek*5 (take ln of each side) Ln .5= ln e5k (use power to product rule) Ln .5 = 5k ln e Ln .5 = 5k k = -.139 now replace P(t) with 15, and k with -.139 and solve for t. 15 = 60e-.139t (divide by 60) .25 = e-..139t (take ln of each side) Ln .25 ln e-.139t Ln .25 = -.139t lne Ln .25 = -.139t Answer: 9.97 years