1. Continuous Growth and the Number e Lesson 3.4

2. Compounding Multiple Times Per Year Given the following formula for compounding  P = initial investment  r = yearly rate  n = number of compounding periods  t = number of years

3. Compounding Multiple Times Per Year What if we invested $1000 for 5 years at 4% interest Try the formula for different numbers of compounding periods  Monthly  Weekly (n = 52)  Daily (n = 365)  Hourly (n = 365 * 25) What phenomenon do you notice?

4. Compounding Multiple Times Per Year You should see that we seem to reach a limit as to how much multiple compounding periods increase the final amount So we come up with continuous compounding

5. Using e As the Base We have used y = A * B t Consider letting B = e k Then by substitution y = A * (e k ) t Recall B = (1 + r) (the growth factor) It turns out that

6. Continuous Growth The constant k is called the continuous percent growth rate For Q = a b t  k can be found by solving e k = b Then Q = a e k*t For positive a  if k > 0 then Q is an increasing function  if k < 0 then Q is a decreasing function

7. Continuous Growth For Q = a e k*t Assume a > 0 k > 0 k < 0

8. Continuous Growth For the function what is the continuous growth rate? The growth rate is the coefficient of t  Growth rate = 0.4 or 40% Graph the function (predict what it looks like)

9. Converting Between Forms Change to the form Q = A*B t We know B = e k Change to the form Q = A*e k*t We will eventually discover that k = ln B

10. Assignment Lesson 3.4 Page 133 Exercises 1 – 25 odd