Section 6.7 – Financial Models Simple Interest Formula 𝐼=𝑃𝑟𝑡 𝐼=𝑎𝑚𝑜𝑢𝑛𝑡 𝑜𝑓 𝑖𝑛𝑡𝑒𝑟𝑒𝑠𝑡 𝑃=𝑝𝑟𝑖𝑛𝑐𝑖𝑝𝑙𝑒 𝑖𝑛𝑣𝑒𝑠𝑡𝑒𝑑 𝑟=𝑎𝑛𝑛𝑢𝑎𝑙 𝑖𝑛𝑡𝑒𝑟𝑒𝑠𝑡 𝑟𝑎𝑡𝑒 𝑒𝑥𝑝𝑟𝑒𝑠𝑠𝑒𝑑 𝑎𝑠 𝑎 𝑑𝑒𝑐𝑖𝑚𝑎𝑙 𝑡=𝑙𝑒𝑛𝑔𝑡ℎ 𝑜𝑓 𝑡𝑖𝑚𝑒 𝑖𝑛 𝑦𝑒𝑎𝑟𝑠 Compound Interest Formula 𝐴=𝑃∙ 1+ 𝑟 𝑛 𝑛∙𝑡 𝐴=𝑟𝑒𝑡𝑢𝑟𝑛 𝑜𝑛 𝑡ℎ𝑒 𝑝𝑟𝑖𝑛𝑐𝑖𝑝𝑙𝑒 𝑛=𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑡𝑖𝑚𝑒𝑠 𝑖𝑛 𝑎 𝑦𝑒𝑎𝑟 𝑖𝑛 𝑤ℎ𝑖𝑐𝑘 𝑖𝑛𝑡𝑒𝑟𝑒𝑠𝑡 𝑖𝑠 𝑐𝑙𝑎𝑢𝑙𝑎𝑡𝑒𝑑 Continuous Compounding Interest Formula 𝐴=𝑃 𝑒 𝑟∙𝑡

Section 6.7 – Financial Models Example - Simple Interest 𝐼=𝑃𝑟𝑡 What is the future value of a $34,100 principle invested at 4% for 3 years 𝐼= 34100 (.04)(3) 𝐹𝑢𝑡𝑢𝑟𝑒 𝑉𝑎𝑙𝑢𝑒=34100+4092.00 𝐼=$4092.00 𝐹𝑢𝑡𝑢𝑟𝑒 𝑉𝑎𝑙𝑢𝑒=$38,192.00 Examples - Compound Interest 𝐴=𝑃∙ 1+ 𝑟 𝑛 𝑛∙𝑡 The amount of $12,700 is invested at 8.8% compounded semiannually for 1 year. What is the future value? 𝐴=12700∙ 1+ 0.088 2 2∙1 𝐴=$13,842.19 $21,000 is invested at 13.6% compounded quarterly for 4 years. What is the return value? 𝐴=21000∙ 1+ 0.136 4 4∙4 𝐴=$35,854.85

Section 6.7 – Financial Models Examples - Compound Interest 𝐴=𝑃∙ 1+ 𝑟 𝑛 𝑛∙𝑡 How much money will you have if you invest $4000 in a bank for sixty years at an annual interest rate of 9%, compounded monthly? 𝐴=4000∙ 1+ 0.09 12 12∙60 𝐴=$867,959.49 Example - Continuous Compounding Interest 𝐴=𝑃 𝑒 𝑟∙𝑡 If you invest $500 at an annual interest rate of 10% compounded continuously, calculate the final amount you will have in the account after five years. 𝐴=500 𝑒 0.10∙5 𝐴=$824.36

Section 6.7 – Financial Models Effective Interest Rate – the actual annual interest rate that takes into account the effects of compounding. Compounding n times per year: 𝑟 𝑒 = 1+ 𝑟 𝑛 𝑛 −1 Continuous compounding: 𝑟 𝑒 = 𝑒 𝑟 −1 Which is better, to receive 9.5% (annual rate) continuously compounded or 10% (annual rate) compounded 4 times per year? Continuous compounding Compounding 4 times per year 𝑟 𝑒 = 𝑒 𝑟 −1 𝑟 𝑒 = 1+ 𝑟 𝑛 𝑛 −1 𝑟 𝑒 = 𝑒 0.095 −1 𝑟 𝑒 = 1+ 0.10 4 4 −1 𝑟 𝑒 =0.1074=10.74% 𝑟 𝑒 =0.1038=10.38%

Section 6.7 – Financial Models Present Value – the initial principal invested at a specific rate and time that will grow to a predetermined value. Compounding n times per year: P=𝐴∙ 1+ 𝑟 𝑛 −𝑛∙𝑡 Continuous compounding: P=𝐴 𝑒 −𝑟∙𝑡 How much money do you have to put in the bank at 12% annual interest for five years (a) compounded 6 times per year and (b) compounded continuously to end up with $2,000? Compounding 6 times per year Continuous compounding P=2000 1+ 0.12 6 −6∙5 𝑃=2000 𝑒 −0.12∙5 P=$1,104.14 𝑃=$1097.62

Section 6.7 – Financial Models Example What rate of interest (a) compounded monthly and (b) continuous compounding is required to triple an investment in five years? 𝐶𝑜𝑚𝑝𝑜𝑢𝑛𝑑𝑒𝑑 𝑀𝑜𝑛𝑡ℎ𝑙𝑦 𝐶𝑜𝑛𝑡𝑖𝑛𝑢𝑜𝑢𝑠 𝐶𝑜𝑚𝑝𝑜𝑢𝑛𝑑𝑒𝑑 𝐴=𝑃∙ 1+ 𝑟 𝑛 𝑛∙𝑡 𝐴=𝑃 𝑒 𝑟∙𝑡 3𝑃=𝑃 𝑒 𝑟∙5 3𝑃=𝑃∙ 1+ 𝑟 12 12∙5 3= 𝑒 𝑟∙5 𝑙𝑛3= 𝑙𝑛𝑒 𝑟∙5 3= 1+ 𝑟 12 60 𝑙𝑛3=5𝑟 𝑟= 𝑙𝑛3 5 60 3 =1+ 𝑟 12 𝑟=0.2197=21.97% 12 60 3 =12+𝑟 12 60 3 −12=𝑟 𝑟=0.2217=22.17%

Section 6.8 – Exponential Growth/Decay Models; Newton's Law of Cooling and Logistic Growth/Decay Models Uninhibited Exponential Growth 𝐴(𝑡)= 𝐴 0 𝑒 𝑘𝑡 𝐴 𝑡 =𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑟 𝑤𝑒𝑖𝑔ℎ𝑡 𝑎𝑓𝑡𝑒𝑟 𝑡𝑖𝑚𝑒 𝐴 0 =𝑖𝑛𝑖𝑡𝑖𝑎𝑙 𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑟 𝑤𝑒𝑖𝑔ℎ𝑡 𝑘=𝑟𝑎𝑡𝑒 𝑜𝑓 𝑔𝑟𝑜𝑤𝑡ℎ; 𝑡ℎ𝑒 𝒑𝒐𝒔𝒊𝒕𝒊𝒗𝒆 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡 (𝑘>0) 𝑡=𝑡𝑖𝑚𝑒 𝑝𝑎𝑠𝑠𝑒𝑑 Uninhibited Exponential Decay 𝐴(𝑡)= 𝐴 0 𝑒 𝑘𝑡 𝐴 𝑡 =𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑟 𝑤𝑒𝑖𝑔ℎ𝑡 𝑎𝑓𝑡𝑒𝑟 𝑡𝑖𝑚𝑒 𝐴 0 =𝑖𝑛𝑖𝑡𝑖𝑎𝑙 𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑟 𝑤𝑒𝑖𝑔ℎ𝑡 𝑘=𝑡ℎ𝑒 𝑟𝑎𝑡𝑒 𝑜𝑓 𝑑𝑒𝑐𝑎𝑦; 𝑡ℎ𝑒 𝒏𝒆𝒈𝒂𝒕𝒊𝒗𝒆 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡 (𝑘<0) 𝑡=𝑡𝑖𝑚𝑒 𝑝𝑎𝑠𝑠𝑒𝑑

Section 6.8 – Exponential Growth/Decay Models; Newton's Law of Cooling and Logistic Growth/Decay Models Examples The population of the United States was approximately 227 million in 1980 and 282 million in 2000. Estimate the population in the years 2010 and 2020. 𝐴(𝑡)= 𝐴 0 𝑒 𝑘𝑡 Find k 2010 𝐴 𝑡 =227 𝑒 0.010848 (2010−1980) 282=227 𝑒 𝑘(2000−1980) 𝐴 𝑡 =314.3 𝑚𝑖𝑙𝑙𝑖𝑜𝑛 282 227 = 𝑒 20𝑘 𝐹𝑟𝑜𝑚 2010 𝐶𝑒𝑛𝑠𝑢𝑠: 308.7 𝑚𝑖𝑙𝑙𝑖𝑜𝑛 𝑙𝑛 282 227 =20𝑘 2020 𝐴 𝑡 =227 𝑒 0.010848 (2020−1980) 𝑘= 𝑙𝑛 282 227 20 =0.010848 𝐴 𝑡 =350.3 𝑚𝑖𝑙𝑙𝑖𝑜𝑛

Section 6.8 – Exponential Growth/Decay Models; Newton's Law of Cooling and Logistic Growth/Decay Models Examples A radioactive material has a half-life of 700 years. If there were ten grams initially, how much would remain after 300 years? When will the material weigh 7.5 grams? 𝐴(𝑡)= 𝐴 0 𝑒 𝑘𝑡 Find k 300 years 7.5 grams 𝐴 𝑡 =10 𝑒 −0.00099 (300) 7.5=10 𝑒 −0.00099 𝑡 1=2 𝑒 𝑘(700) 𝐴 𝑡 =7.43 𝑔𝑟𝑎𝑚𝑠 0.75= 𝑒 −0.00099 𝑡 0.5= 𝑒 700𝑘 or 𝑙𝑛0.75=−0.00099𝑡 𝑙𝑛0.5=700𝑘 𝑘= 𝑙𝑛0.5 700 𝐴 𝑡 =10 𝑒 𝑙𝑛0.5 700 (300) 𝑡=290.6 𝑦𝑒𝑎𝑟𝑠 or 𝑘=−0.000990 𝐴 𝑡 =7.43 𝑔𝑟𝑎𝑚𝑠 7.5=10 𝑒 𝑙𝑛0.5 700 𝑡 0.75= 𝑒 𝑙𝑛0.5 700 𝑡 𝑙𝑛0.75= 𝑙𝑛0.5 700 𝑡 𝑡=290.5 𝑦𝑒𝑎𝑟𝑠

Section 6.8 – Exponential Growth/Decay Models; Newton's Law of Cooling and Logistic Growth/Decay Models Newton’s Law of Cooling 𝑢 𝑡 =𝑇+( 𝑢 0 −𝑇) 𝑒 𝑘𝑡 𝑢 𝑡 =𝑡𝑒𝑚𝑝𝑒𝑟𝑎𝑡𝑢𝑟𝑒 𝑜𝑓 𝑎 ℎ𝑒𝑎𝑡𝑒𝑑 𝑜𝑏𝑗𝑒𝑐𝑡 𝑎𝑡 𝑎𝑛𝑦 𝑔𝑖𝑣𝑒𝑛 𝑡𝑖𝑚𝑒 𝑇=𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡 𝑡𝑒𝑚𝑝𝑒𝑟𝑎𝑡𝑢𝑟𝑒 𝑜𝑓 𝑡ℎ𝑒 𝑠𝑢𝑟𝑟𝑜𝑢𝑛𝑑𝑖𝑛𝑔 𝑚𝑒𝑑𝑖𝑢𝑚 𝑢 0 =𝑖𝑛𝑖𝑡𝑖𝑎𝑙 𝑡𝑒𝑚𝑝𝑒𝑟𝑎𝑡𝑢𝑟𝑒 𝑜𝑓 𝑡ℎ𝑒 ℎ𝑒𝑎𝑡𝑒𝑑 𝑜𝑏𝑗𝑒𝑐𝑡 𝑘=𝑛𝑒𝑔𝑎𝑡𝑖𝑣𝑒 𝑐𝑜𝑜𝑙𝑖𝑛𝑔 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡 (𝑘<0) 𝑡=𝑡𝑖𝑚𝑒 𝑝𝑎𝑠𝑠𝑒𝑑

Section 6.8 – Exponential Growth/Decay Models; Newton's Law of Cooling and Logistic Growth/Decay Models Newton’s Law of Cooling 𝑢 𝑡 =𝑇+( 𝑢 0 −𝑇) 𝑒 𝑘𝑡 Example A pizza pan is removed at 3:00 PM from an oven whose temperature is fixed at 450 F into a room that is a constant 70 F. After 5 minutes, the pizza pan is at 300 F. At what time is the temperature of the pan 135 F? Find k t @ 135 F 300=70+(450−70) 𝑒 𝑘5 135=70+(450−70) 𝑒 −0.100418𝑡 230=380 𝑒 𝑘5 65=380 𝑒 −0.100418𝑡 230 380 = 𝑒 𝑘5 65 380 = 𝑒 −0.100418𝑡 𝑙𝑛 23 38 =5𝑘 𝑙𝑛 13 75 =−0.100418𝑡 𝑘= 𝑙𝑛 23 38 5 =−0.100418 𝑡=17.45 𝑚𝑖𝑛𝑢𝑡𝑒𝑠 𝑎𝑏𝑜𝑢𝑡 3:17 𝑃𝑀

Section 6.8 – Exponential Growth/Decay Models; Newton's Law of Cooling and Logistic Growth/Decay Models Logistic Growth/Decay 𝑃 𝑡 = 𝑐 1+𝑎 𝑒 −𝑏𝑡 𝑃 𝑡 =𝑝𝑜𝑝𝑢𝑙𝑎𝑡𝑖𝑜𝑛 𝑎𝑓𝑡𝑒𝑟 𝑡𝑖𝑚𝑒 𝑎, 𝑏, 𝑐=𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡𝑠 𝑜𝑏𝑡𝑎𝑖𝑛𝑒𝑑 𝑡ℎ𝑟𝑜𝑢𝑔ℎ 𝑑𝑎𝑡𝑎 𝑎𝑛𝑎𝑙𝑦𝑠𝑖𝑠 (𝑎>0 𝑎𝑛𝑑 𝑐>0) 𝑔𝑟𝑜𝑤𝑡ℎ 𝑚𝑜𝑑𝑒𝑙 𝑖𝑓 𝑏>0 𝑑𝑒𝑐𝑎𝑦 𝑚𝑜𝑑𝑒𝑙 𝑖𝑓 𝑏<0 𝑡=𝑡𝑖𝑚𝑒 𝑝𝑎𝑠𝑠𝑒𝑑

Section 6.8 – Exponential Growth/Decay Models; Newton's Law of Cooling and Logistic Growth/Decay Models 𝑃 𝑡 = 𝑐 1+𝑎 𝑒 −𝑏𝑡 Logistic Growth/Decay Example The logistic growth model 𝑃 𝑡 = 0.95 1+3 𝑒 −0.32𝑡 relates the proportion of U.S. households that own a cell phone to the year. Let 𝑡=0 represent 2000, 𝑡=1 represent 2001, and so on. (a) What proportion of households owned a cell phone in 2000, (b) what proportion of households owned a cell phone in 2005, and (c) when will 85% of the households own a cell phone? 2000 2005 P(t) = 85% 𝑡=2000−2000=0 𝑡=2005−2000=5 0.85= 0.95 1+3 𝑒 −0.32𝑡 𝑃 𝑡 = 0.95 1+3 𝑒 −0.32(0) 𝑃 𝑡 = 0.95 1+3 𝑒 −0.32(5) 0.85(1+3 𝑒 −0.32𝑡 )=0.95 𝑃 𝑡 = 0.95 1+3 𝑃 𝑡 = 0.95 1.60569 0.85+2.55 𝑒 −0.32𝑡 =0.95 2.55 𝑒 −0.32𝑡 =0.10 𝑃 𝑡 =0.2375=23.75% 𝑃 𝑡 =0.5916=59.16% 𝑒 −0.32𝑡 =0.039216 −0.32𝑡=𝑙𝑛0.039216 𝑡=10.12 𝑦𝑒𝑎𝑟𝑠 →2010 𝑡𝑜 2011