The growth of a population depends on its initial size.
Delta Notation: Greek letter D = Change in quantity
Rate of Change =
Average growth rate:
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.
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
= 0.06666666 = 6.7%
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.
Use data from example 1.
t = time in years
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
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,
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