Exponential Growth The growth of a population depends on its initial size. Delta Notation: Greek letter D = Change in quantity Rate of Change =
one change Average growth rate: Example 1: In 1980, Anytown USA had a population of 20,000. Assume change in population to births and deaths only. In 1981, population was 20,600. What is the average growth rate? D Pop = 20,600 20,000 = 600 people D Time = 1981 1980 = 1 year What is the likely average growth rate for 1982? 600/20,000 = .03 = 3% If the population continues at the same rate, how many more people are there in 1982? X/20,600 = .03 X = 618 So the population in 1982 is 21,218. Example 2: A house bought for $150,000 in Jan. 1995 was appraised for $160,000 a year later. What would the house be worth in 1997 assuming the same rate of appreciation. D Value = $160,000 150,000 = $10,000 D Time = 1996 1995 = 1 year Rate of appreciation = $10,000/year
10,000
= 0.06666666 = 6.7%
x
=
.066666666 Exponential Growth Model Our last Example was tedious and did not offer a model to predict any years appreciation value. e is constant = 2.718 . b is rate y is resulting value or population This growth model works for all forms of growth, population, real estate appreciation etc. With a few data, we can predict many aspects of the equation. Developing individual models. Example 3: Use data from example 1. Using t = time in years and p = population (t, p) = (0, 20,000) and (1, 20,600) we get: 20,000 = ae^{b(0)} = a = 20,000 our initial population 20,600 = 20,000e^{b(1)} isolate e by dividing both sides by 20,000 1.03 = e^{b} take ln of both sides ln 1.03 = ln e^{b} = b inverse property b = 0.029558802 the rate of growth Notice that it is almost the average growth rate 3% What is the population in 10 years? y = 20,000e^{.029558802(10)} = 26,878 people When will the population double? y = 40,000 if doubled 40,000 = 20,000e^{.029558802t} isolate e 2 = e^{.029558802t} take ln of both sides ln 2 = ln e^{.029558802t} =.029558802t solve for t ln 2/0.029558802 = t = 23.44 years Example 4: Find a growth model for the real estate appreciation in Example 2. 1995 = time 0 (t, v) = (0, $150,000) and (1, 160,000) Since we know that at t=0, v= $150,000, then a = initial amount = $150,000 $160,000 = $150,000e^{b(1)} isolate e 1.06= e^{b }take ln of both sides ln 1.06= ln e^{b} = b inverse property b = 0.064538521138 rate of appreciation what will the house be worth in the year 2000? T = 5 y = 150,000e^{.064538521138(5) } = $207,126.12
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