Exponential Growth Section 9.1   The growth of a population depends on its initial size. Delta Notation: Greek letter     D     =     Change in quantity Rate of Change =      one change                                 Another 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           Average Growth rate = 600 people/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%          150,000             x           = .066666666         160,000           x = 10,666.67           \$160,000 + 10,666.67 = \$170,666.67 in 1997 Exponential Growth Model Our last Example was tedious and did not offer a model to predict any years appreciation value.           y = aebt           where a is initial amount, value or population,                                     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 = aeb(0) = a = 20,000            our initial population         20,600 = 20,000eb(1)                         isolate e by dividing both sides by 20,000         1.03 = eb                                            take ln of both sides         ln 1.03 = ln eb = 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 = \$150,000eb(1)                            isolate e         1.06= eb                                                                          take ln of both sides         ln 1.06= ln eb = 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   Back to Exponenials and Logarithms Main Page Back to the Survey of Math Ideas Home Page e-mail Questions and Suggestions