1. Warm Up 100 grams of a compound with a half-life of 5000 years
Write a function to represent the amount after t years for each situation.
100 grams of a compound with a half-life of 5000 years
12 bacteria that quadruple themselves every 2 years
A new car worth $35,000 that depreciates 15% per year
A $75,000 student loan with a 6% annual interest rate
𝑨 𝒕 =𝟏𝟎𝟎 ( 𝟏 𝟐 ) 𝒕 𝟓𝟎𝟎𝟎
𝑨 𝒕 =𝟏𝟐 (𝟒) 𝒕 𝟐
𝑨 𝒕 =𝟑𝟓,𝟎𝟎𝟎 (𝟏−𝟎.𝟏𝟓) 𝒕 =𝟑𝟓𝟎𝟎𝟎 (𝟎.𝟖𝟓) 𝒕
𝑨 𝒕 =𝟕𝟓,𝟎𝟎𝟎 (𝟏+𝟎.𝟎𝟔) 𝒕 =𝟕𝟓,𝟎𝟎𝟎 (𝟏.𝟎𝟔) 𝒕

2. Exponential Equation Reminders
𝒇(𝒙)=𝒂 𝒃 𝒙
a= starting value b= multiplier
𝑨 𝒕 = 𝑨 𝟎 (𝟏+𝒓) 𝒕
Exponential growth and decay- given a rate
r is the rate, as a decimal
r is positive for growth, negative for decay
t is positive for the future, negative for the past
𝑨 𝒕 = 𝑨 𝟎 (𝒃) 𝒕 𝒌 or 𝑨 𝟎 𝒃 𝟏 𝒌 𝒕
Exponential growth and decay- given an outcome
and the time to achieve it
b is the outcome, in other words, “what happened”
examples: reduced by ½, doubled, multiplied by 4
k is how long it takes to do that

3. 5.4 The Function 𝑒 𝑥 Complete the table. Lim 10 2.5937 100 2.7048 1000
2.7169
10,000
2.7181
100,000
2.7183
Lim

4. 5.4 The Function 𝑒 𝑥

5. Leonhard Euler http://www.storyofmathematics.com/18th_e uler.html

6. Facts About Leonard Euler
He spent most of his academic life in Russia and Germany.
He had a long life and thirteen children.
His collected works comprise nearly 900 books and, in the year 1775, he is said to have produced on average one mathematical paper every week. 
He had a photographic memory. 

8. Compound Interest Equation
𝑷 𝒕 = 𝑷 𝟎 𝟏+ 𝒓 𝒌 𝒕𝒌
𝑃 0 = initial investment
r = annual rate as a decimal
k = number of times compounded per year
t = time

9. 5.4 The Function e^x

10. Compound Interest Equation
𝑷 𝒕 = 𝑷 𝟎 𝟏+ 𝒓 𝒌 𝒕𝒌
𝑃 0 = initial investment
r = rate as a decimal
k = number of times compounded per year
t = time
If you invest $500 at a 10% annual interest rate that is compounded monthly, how much money will you have after 6 years?
𝑃 6 = ∙12 =$908.80

11. Compound Interest

12. % growth each period Amount Annually 12% 12%÷2=6% Quarterly 12%÷4=3%
12% Annual Growth
12% Annual Growth
Compounded:
% growth each period
Growth factor
during period
Amount
Annually
12%
Semiannually
12%÷2=6%
Quarterly
12%÷4=3%
Monthly
12%÷12=1%
Daily
(365 days)
(12/365)%
k times per year
(12/k)%

13. Lim

14. 𝑃 6 =500 𝑒 .1∙6 =$911.06 𝑃 0 = initial investment
r = rate as a decimal t = time
If you invest $500 at a 10% annual interest rate that is compounded continuously, how much money will you have after 6 years?
𝑃 6 =500 𝑒 .1∙6 =$911.06

15. Effective Annual Yield
Actual % growth after 1 year
This will be more than the given annual rate, if interest is compounded more than once per year.
𝑎𝑚𝑜𝑢𝑛𝑡 𝑎𝑓𝑡𝑒𝑟 1 𝑦𝑒𝑎𝑟−𝑜𝑟𝑖𝑔𝑖𝑛𝑎𝑙 𝑖𝑛𝑣𝑒𝑠𝑡𝑚𝑒𝑛𝑡 𝑜𝑟𝑖𝑔𝑖𝑛𝑎𝑙 𝑖𝑛𝑣𝑒𝑠𝑡𝑚𝑒𝑛𝑡 ∗100
After a year during which interest is compounded quarterly, an investment of $750 is worth $790. What is the effective annual yield?
790− ∗100=5.33%

16. Homework
Page 189 # 1-11 odd, all