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

 

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