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Continuous Growth and the Number e

Continuous Growth and the Number e. Lesson 3.4. 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. Compounding Multiple Times Per Year.

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Continuous Growth and the Number e

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  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 yearsat 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 * Bt • Consider letting B = ek • Then by substitution y = A * (ek)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 bt • k can be found by solving ek = b • Then Q = a ek*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 ek*t Assume a > 0 • k > 0 • k < 0

  8. Continuous Growth • For the functionwhat is thecontinuous 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*Bt • We know B = ek • Change to the form Q = A*ek*t • We will eventually discover that k = ln B

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

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