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Published byLionel Alvin Short Modified over 9 years ago
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Continuous Growth and the Number e Lesson 3.4
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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
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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?
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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
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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
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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
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Continuous Growth For Q = a e k*t Assume a > 0 k > 0 k < 0
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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)
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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
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Assignment Lesson 3.4 Page 133 Exercises 1 – 25 odd
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