1. 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

2. Section 6.7 – Financial Models
Example - Simple Interest
𝐼=𝑃𝑟𝑡
What is the future value of a $34,100 principle invested at 4% for 3 years
𝐼= (.04)(3)
𝐹𝑢𝑡𝑢𝑟𝑒 𝑉𝑎𝑙𝑢𝑒=
𝐼=$
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
𝐴=$13,842.19
$21,000 is invested at 13.6% compounded quarterly for 4 years. What is the return value?
𝐴=21000∙ ∙4
𝐴=$35,854.85

3. 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∙ ∙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

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: 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 −1
𝑟 𝑒 = −1
𝑟 𝑒 =0.1074=10.74%
𝑟 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=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= −6∙5
𝑃=2000 𝑒 −0.12∙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?
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 ∙5
3= 𝑒 𝑟∙5
𝑙𝑛3= 𝑙𝑛𝑒 𝑟∙5
3= 1+ 𝑟
𝑙𝑛3=5𝑟
𝑟= 𝑙𝑛3 5
60 3 =1+ 𝑟 12
𝑟=0.2197=21.97%
=12+𝑟
−12=𝑟
𝑟=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(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

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(𝑡)= 𝐴 0 𝑒 𝑘𝑡
Find k
2010
𝐴 𝑡 =227 𝑒 (2010−1980)
282=227 𝑒 𝑘(2000−1980)
𝐴 𝑡 =314.3 𝑚𝑖𝑙𝑙𝑖𝑜𝑛
= 𝑒 20𝑘
𝐹𝑟𝑜𝑚 2010 𝐶𝑒𝑛𝑠𝑢𝑠: 𝑚𝑖𝑙𝑙𝑖𝑜𝑛
𝑙𝑛 =20𝑘
2020
𝐴 𝑡 =227 𝑒 (2020−1980)
𝑘= 𝑙𝑛 =
𝐴 𝑡 =350.3 𝑚𝑖𝑙𝑙𝑖𝑜𝑛

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(𝑡)= 𝐴 0 𝑒 𝑘𝑡
Find k
300 years
7.5 grams
𝐴 𝑡 =10 𝑒 − (300)
7.5=10 𝑒 − 𝑡
1=2 𝑒 𝑘(700)
𝐴 𝑡 =7.43 𝑔𝑟𝑎𝑚𝑠
0.75= 𝑒 − 𝑡
0.5= 𝑒 700𝑘 or
𝑙𝑛0.75=− 𝑡
𝑙𝑛0.5=700𝑘
𝑘= 𝑙𝑛
𝐴 𝑡 =10 𝑒 𝑙𝑛 (300)
𝑡=290.6 𝑦𝑒𝑎𝑟𝑠 or
𝑘=−
𝐴 𝑡 =7.43 𝑔𝑟𝑎𝑚𝑠
7.5=10 𝑒 𝑙𝑛 𝑡
0.75= 𝑒 𝑙𝑛 𝑡
𝑙𝑛0.75= 𝑙𝑛 𝑡
𝑡=290.5 𝑦𝑒𝑎𝑟𝑠

10. 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

11. 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
135 F
300=70+(450−70) 𝑒 𝑘5
135=70+(450−70) 𝑒 − 𝑡
230=380 𝑒 𝑘5
65=380 𝑒 − 𝑡
= 𝑒 𝑘5
= 𝑒 − 𝑡
𝑙𝑛 =5𝑘
𝑙𝑛 =− 𝑡
𝑘= 𝑙𝑛 =−
𝑡=17.45 𝑚𝑖𝑛𝑢𝑡𝑒𝑠
𝑎𝑏𝑜𝑢𝑡 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 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

13. 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 = 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= 𝑒 −0.32𝑡
𝑃 𝑡 = 𝑒 −0.32(0)
𝑃 𝑡 = 𝑒 −0.32(5)
0.85(1+3 𝑒 −0.32𝑡 )=0.95
𝑃 𝑡 =
𝑃 𝑡 =
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 =
−0.32𝑡=𝑙𝑛
𝑡=10.12 𝑦𝑒𝑎𝑟𝑠
→2010 𝑡𝑜 2011