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Section 6.7 – Financial Models

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1 Section 6.7 – Financial Models
Simple Interest Formula 𝐼=π‘ƒπ‘Ÿπ‘‘ 𝐼=π‘Žπ‘šπ‘œπ‘’π‘›π‘‘ π‘œπ‘“ π‘–π‘›π‘‘π‘’π‘Ÿπ‘’π‘ π‘‘ 𝑃=π‘π‘Ÿπ‘–π‘›π‘π‘–π‘π‘™π‘’ 𝑖𝑛𝑣𝑒𝑠𝑑𝑒𝑑 π‘Ÿ=π‘Žπ‘›π‘›π‘’π‘Žπ‘™ π‘–π‘›π‘‘π‘’π‘Ÿπ‘’π‘ π‘‘ π‘Ÿπ‘Žπ‘‘π‘’ 𝑒π‘₯π‘π‘Ÿπ‘’π‘ π‘ π‘’π‘‘ π‘Žπ‘  π‘Ž π‘‘π‘’π‘π‘–π‘šπ‘Žπ‘™ 𝑑=π‘™π‘’π‘›π‘”π‘‘β„Ž π‘œπ‘“ π‘‘π‘–π‘šπ‘’ 𝑖𝑛 π‘¦π‘’π‘Žπ‘Ÿπ‘  Compound Interest Formula 𝐴=π‘ƒβˆ™ 1+ π‘Ÿ 𝑛 π‘›βˆ™π‘‘ 𝐴=π‘Ÿπ‘’π‘‘π‘’π‘Ÿπ‘› π‘œπ‘› π‘‘β„Žπ‘’ π‘π‘Ÿπ‘–π‘›π‘π‘–π‘π‘™π‘’ 𝑛=π‘›π‘’π‘šπ‘π‘’π‘Ÿ π‘œπ‘“ π‘‘π‘–π‘šπ‘’π‘  𝑖𝑛 π‘Ž π‘¦π‘’π‘Žπ‘Ÿ 𝑖𝑛 π‘€β„Žπ‘–π‘π‘˜ π‘–π‘›π‘‘π‘’π‘Ÿπ‘’π‘ π‘‘ 𝑖𝑠 π‘π‘™π‘Žπ‘’π‘™π‘Žπ‘‘π‘’π‘‘ Continuous Compounding Interest Formula 𝐴=𝑃 𝑒 π‘Ÿβˆ™π‘‘

2 Section 6.7 – Financial Models
Example - Simple Interest 𝐼=π‘ƒπ‘Ÿπ‘‘ What is the future value of a $34,100 principle invested at 4% for 3 years 𝐼= (.04)(3) πΉπ‘’π‘‘π‘’π‘Ÿπ‘’ π‘‰π‘Žπ‘™π‘’π‘’= 𝐼=$ πΉπ‘’π‘‘π‘’π‘Ÿπ‘’ π‘‰π‘Žπ‘™π‘’π‘’=$38,192.00 Examples - Compound Interest 𝐴=π‘ƒβˆ™ 1+ π‘Ÿ 𝑛 π‘›βˆ™π‘‘ The amount of $12,700 is invested at 8.8% compounded semiannually for 1 year. What is the future value? 𝐴=12700βˆ™ βˆ™1 𝐴=$13,842.19 $21,000 is invested at 13.6% compounded quarterly for 4 years. What is the return value? 𝐴=21000βˆ™ βˆ™4 𝐴=$35,854.85

3 Section 6.7 – Financial Models
Examples - Compound Interest 𝐴=π‘ƒβˆ™ 1+ π‘Ÿ 𝑛 π‘›βˆ™π‘‘ 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=4000βˆ™ βˆ™60 𝐴=$867,959.49 Example - Continuous Compounding Interest 𝐴=𝑃 𝑒 π‘Ÿβˆ™π‘‘ 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 0.10βˆ™5 𝐴=$824.36

4 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: π‘Ÿ 𝑒 = 1+ π‘Ÿ 𝑛 𝑛 βˆ’1 Continuous compounding: π‘Ÿ 𝑒 = 𝑒 π‘Ÿ βˆ’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 π‘Ÿ 𝑒 = 𝑒 π‘Ÿ βˆ’1 π‘Ÿ 𝑒 = 1+ π‘Ÿ 𝑛 𝑛 βˆ’1 π‘Ÿ 𝑒 = 𝑒 βˆ’1 π‘Ÿ 𝑒 = βˆ’1 π‘Ÿ 𝑒 =0.1074=10.74% π‘Ÿ 𝑒 =0.1038=10.38%

5 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=π΄βˆ™ 1+ π‘Ÿ 𝑛 βˆ’π‘›βˆ™π‘‘ Continuous compounding: P=𝐴 𝑒 βˆ’π‘Ÿβˆ™π‘‘ 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= βˆ’6βˆ™5 𝑃=2000 𝑒 βˆ’0.12βˆ™5 P=$1,104.14 𝑃=$

6 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? πΆπ‘œπ‘šπ‘π‘œπ‘’π‘›π‘‘π‘’π‘‘ π‘€π‘œπ‘›π‘‘β„Žπ‘™π‘¦ πΆπ‘œπ‘›π‘‘π‘–π‘›π‘’π‘œπ‘’π‘  πΆπ‘œπ‘šπ‘π‘œπ‘’π‘›π‘‘π‘’π‘‘ 𝐴=π‘ƒβˆ™ 1+ π‘Ÿ 𝑛 π‘›βˆ™π‘‘ 𝐴=𝑃 𝑒 π‘Ÿβˆ™π‘‘ 3𝑃=𝑃 𝑒 π‘Ÿβˆ™5 3𝑃=π‘ƒβˆ™ 1+ π‘Ÿ βˆ™5 3= 𝑒 π‘Ÿβˆ™5 𝑙𝑛3= 𝑙𝑛𝑒 π‘Ÿβˆ™5 3= 1+ π‘Ÿ 𝑙𝑛3=5π‘Ÿ π‘Ÿ= 𝑙𝑛3 5 60 3 =1+ π‘Ÿ 12 π‘Ÿ=0.2197=21.97% =12+π‘Ÿ βˆ’12=π‘Ÿ π‘Ÿ=0.2217=22.17%

7 Section 6.8 – Exponential Growth/Decay Models;
Newton's Law of Cooling and Logistic Growth/Decay Models Uninhibited Exponential Growth 𝐴(𝑑)= 𝐴 0 𝑒 π‘˜π‘‘ 𝐴 𝑑 =π‘›π‘’π‘šπ‘π‘’π‘Ÿ π‘œπ‘Ÿ π‘€π‘’π‘–π‘”β„Žπ‘‘ π‘Žπ‘“π‘‘π‘’π‘Ÿ π‘‘π‘–π‘šπ‘’ 𝐴 0 =π‘–π‘›π‘–π‘‘π‘–π‘Žπ‘™ π‘›π‘’π‘šπ‘π‘’π‘Ÿ π‘œπ‘Ÿ π‘€π‘’π‘–π‘”β„Žπ‘‘ π‘˜=π‘Ÿπ‘Žπ‘‘π‘’ π‘œπ‘“ π‘”π‘Ÿπ‘œπ‘€π‘‘β„Ž; π‘‘β„Žπ‘’ π’‘π’π’”π’Šπ’•π’Šπ’—π’† π‘π‘œπ‘›π‘ π‘‘π‘Žπ‘›π‘‘ (π‘˜>0) 𝑑=π‘‘π‘–π‘šπ‘’ π‘π‘Žπ‘ π‘ π‘’π‘‘ Uninhibited Exponential Decay 𝐴(𝑑)= 𝐴 0 𝑒 π‘˜π‘‘ 𝐴 𝑑 =π‘›π‘’π‘šπ‘π‘’π‘Ÿ π‘œπ‘Ÿ π‘€π‘’π‘–π‘”β„Žπ‘‘ π‘Žπ‘“π‘‘π‘’π‘Ÿ π‘‘π‘–π‘šπ‘’ 𝐴 0 =π‘–π‘›π‘–π‘‘π‘–π‘Žπ‘™ π‘›π‘’π‘šπ‘π‘’π‘Ÿ π‘œπ‘Ÿ π‘€π‘’π‘–π‘”β„Žπ‘‘ π‘˜=π‘‘β„Žπ‘’ π‘Ÿπ‘Žπ‘‘π‘’ π‘œπ‘“ π‘‘π‘’π‘π‘Žπ‘¦; π‘‘β„Žπ‘’ π’π’†π’ˆπ’‚π’•π’Šπ’—π’† π‘π‘œπ‘›π‘ π‘‘π‘Žπ‘›π‘‘ (π‘˜<0) 𝑑=π‘‘π‘–π‘šπ‘’ π‘π‘Žπ‘ π‘ π‘’π‘‘

8 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 Estimate the population in the years 2010 and 2020. 𝐴(𝑑)= 𝐴 0 𝑒 π‘˜π‘‘ Find k 2010 𝐴 𝑑 =227 𝑒 (2010βˆ’1980) 282=227 𝑒 π‘˜(2000βˆ’1980) 𝐴 𝑑 =314.3 π‘šπ‘–π‘™π‘™π‘–π‘œπ‘› = 𝑒 20π‘˜ πΉπ‘Ÿπ‘œπ‘š 2010 𝐢𝑒𝑛𝑠𝑒𝑠: π‘šπ‘–π‘™π‘™π‘–π‘œπ‘› 𝑙𝑛 =20π‘˜ 2020 𝐴 𝑑 =227 𝑒 (2020βˆ’1980) π‘˜= 𝑙𝑛 = 𝐴 𝑑 =350.3 π‘šπ‘–π‘™π‘™π‘–π‘œπ‘›

9 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(𝑑)= 𝐴 0 𝑒 π‘˜π‘‘ Find k 300 years 7.5 grams 𝐴 𝑑 =10 𝑒 βˆ’ (300) 7.5=10 𝑒 βˆ’ 𝑑 1=2 𝑒 π‘˜(700) 𝐴 𝑑 =7.43 π‘”π‘Ÿπ‘Žπ‘šπ‘  0.75= 𝑒 βˆ’ 𝑑 0.5= 𝑒 700π‘˜ or 𝑙𝑛0.75=βˆ’ 𝑑 𝑙𝑛0.5=700π‘˜ π‘˜= 𝑙𝑛 𝐴 𝑑 =10 𝑒 𝑙𝑛 (300) 𝑑=290.6 π‘¦π‘’π‘Žπ‘Ÿπ‘  or π‘˜=βˆ’ 𝐴 𝑑 =7.43 π‘”π‘Ÿπ‘Žπ‘šπ‘  7.5=10 𝑒 𝑙𝑛 𝑑 0.75= 𝑒 𝑙𝑛 𝑑 𝑙𝑛0.75= 𝑙𝑛 𝑑 𝑑=290.5 π‘¦π‘’π‘Žπ‘Ÿπ‘ 

10 Section 6.8 – Exponential Growth/Decay Models;
Newton's Law of Cooling and Logistic Growth/Decay Models Newton’s Law of Cooling 𝑒 𝑑 =𝑇+( 𝑒 0 βˆ’π‘‡) 𝑒 π‘˜π‘‘ 𝑒 𝑑 =π‘‘π‘’π‘šπ‘π‘’π‘Ÿπ‘Žπ‘‘π‘’π‘Ÿπ‘’ π‘œπ‘“ π‘Ž β„Žπ‘’π‘Žπ‘‘π‘’π‘‘ π‘œπ‘π‘—π‘’π‘π‘‘ π‘Žπ‘‘ π‘Žπ‘›π‘¦ 𝑔𝑖𝑣𝑒𝑛 π‘‘π‘–π‘šπ‘’ 𝑇=π‘π‘œπ‘›π‘ π‘‘π‘Žπ‘›π‘‘ π‘‘π‘’π‘šπ‘π‘’π‘Ÿπ‘Žπ‘‘π‘’π‘Ÿπ‘’ π‘œπ‘“ π‘‘β„Žπ‘’ π‘ π‘’π‘Ÿπ‘Ÿπ‘œπ‘’π‘›π‘‘π‘–π‘›π‘” π‘šπ‘’π‘‘π‘–π‘’π‘š 𝑒 0 =π‘–π‘›π‘–π‘‘π‘–π‘Žπ‘™ π‘‘π‘’π‘šπ‘π‘’π‘Ÿπ‘Žπ‘‘π‘’π‘Ÿπ‘’ π‘œπ‘“ π‘‘β„Žπ‘’ β„Žπ‘’π‘Žπ‘‘π‘’π‘‘ π‘œπ‘π‘—π‘’π‘π‘‘ π‘˜=π‘›π‘’π‘”π‘Žπ‘‘π‘–π‘£π‘’ π‘π‘œπ‘œπ‘™π‘–π‘›π‘” π‘π‘œπ‘›π‘ π‘‘π‘Žπ‘›π‘‘ (π‘˜<0) 𝑑=π‘‘π‘–π‘šπ‘’ π‘π‘Žπ‘ π‘ π‘’π‘‘

11 Section 6.8 – Exponential Growth/Decay Models;
Newton's Law of Cooling and Logistic Growth/Decay Models Newton’s Law of Cooling 𝑒 𝑑 =𝑇+( 𝑒 0 βˆ’π‘‡) 𝑒 π‘˜π‘‘ Example A pizza pan is removed at 3:00 PM from an oven whose temperature is fixed at 450ο‚° F into a room that is a constant 70ο‚° F. After 5 minutes, the pizza pan is at 300ο‚° F. At what time is the temperature of the pan 135ο‚° F? Find k 135ο‚° F 300=70+(450βˆ’70) 𝑒 π‘˜5 135=70+(450βˆ’70) 𝑒 βˆ’ 𝑑 230=380 𝑒 π‘˜5 65=380 𝑒 βˆ’ 𝑑 = 𝑒 π‘˜5 = 𝑒 βˆ’ 𝑑 𝑙𝑛 =5π‘˜ 𝑙𝑛 =βˆ’ 𝑑 π‘˜= 𝑙𝑛 =βˆ’ 𝑑=17.45 π‘šπ‘–π‘›π‘’π‘‘π‘’π‘  π‘Žπ‘π‘œπ‘’π‘‘ 3:17 𝑃𝑀

12 Section 6.8 – Exponential Growth/Decay Models;
Newton's Law of Cooling and Logistic Growth/Decay Models Logistic Growth/Decay 𝑃 𝑑 = 𝑐 1+π‘Ž 𝑒 βˆ’π‘π‘‘ 𝑃 𝑑 =π‘π‘œπ‘π‘’π‘™π‘Žπ‘‘π‘–π‘œπ‘› π‘Žπ‘“π‘‘π‘’π‘Ÿ π‘‘π‘–π‘šπ‘’ π‘Ž, 𝑏, 𝑐=π‘π‘œπ‘›π‘ π‘‘π‘Žπ‘›π‘‘π‘  π‘œπ‘π‘‘π‘Žπ‘–π‘›π‘’π‘‘ π‘‘β„Žπ‘Ÿπ‘œπ‘’π‘”β„Ž π‘‘π‘Žπ‘‘π‘Ž π‘Žπ‘›π‘Žπ‘™π‘¦π‘ π‘–π‘  (π‘Ž>0 π‘Žπ‘›π‘‘ 𝑐>0) π‘”π‘Ÿπ‘œπ‘€π‘‘β„Ž π‘šπ‘œπ‘‘π‘’π‘™ 𝑖𝑓 𝑏>0 π‘‘π‘’π‘π‘Žπ‘¦ π‘šπ‘œπ‘‘π‘’π‘™ 𝑖𝑓 𝑏<0 𝑑=π‘‘π‘–π‘šπ‘’ π‘π‘Žπ‘ π‘ π‘’π‘‘

13 Section 6.8 – Exponential Growth/Decay Models;
Newton's Law of Cooling and Logistic Growth/Decay Models 𝑃 𝑑 = 𝑐 1+π‘Ž 𝑒 βˆ’π‘π‘‘ Logistic Growth/Decay Example The logistic growth model 𝑃 𝑑 = 𝑒 βˆ’0.32𝑑 relates the proportion of U.S. households that own a cell phone to the year. Let 𝑑=0 represent 2000, 𝑑=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% 𝑑=2000βˆ’2000=0 𝑑=2005βˆ’2000=5 0.85= 𝑒 βˆ’0.32𝑑 𝑃 𝑑 = 𝑒 βˆ’0.32(0) 𝑃 𝑑 = 𝑒 βˆ’0.32(5) 0.85(1+3 𝑒 βˆ’0.32𝑑 )=0.95 𝑃 𝑑 = 𝑃 𝑑 = 𝑒 βˆ’0.32𝑑 =0.95 2.55 𝑒 βˆ’0.32𝑑 =0.10 𝑃 𝑑 =0.2375=23.75% 𝑃 𝑑 =0.5916=59.16% 𝑒 βˆ’0.32𝑑 = βˆ’0.32𝑑=𝑙𝑛 𝑑=10.12 π‘¦π‘’π‘Žπ‘Ÿπ‘  β†’2010 π‘‘π‘œ 2011


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