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Chapter 4: Growth
Population growth is a current topic in the media today. The world population is growing by over 70 million people every year. Predicting populations in the future can have an impact on how countries plan to manage resources for more people. The tools needed to help make predictions about future populations are growth models like the exponential function. This chapter will discuss real world phenomena, like population growth and radioactive decay, using three different growth models.
The growth functions to be examined are linear, exponential, and logistic growth models. Each type of model will be used when data behaves in a specific way and for different types of scenarios. Data that grows by the same amount in each iteration uses a different model than data that increases by a percentage.
Section 4.1: Linear Growth
Starting at the age of 25, imagine if you could save $20 per week, every week, until you retire, how much money would you have stuffed under your mattress at age 65? To solve this problem, we could use a linear growth model. Linear growth has the characteristic of growing by the same amount in each unit of time. In this example, there is an increase of $20 per week; a constant amount is placed under the mattress in the same unit of time.
If we start with $0 under the mattress, then at the end of the first year we would have $20 × 52 = $1040. So, this means you could add $1040 under your mattress every year. At the end of 40 years, you would have $1040 × 40 = $41, 600for retirement. This is not the
best way to save money, but we can see that it is calculated in a systematic way.
Linear Growth: A quantity grows linearly if it grows by a constant amount for each unit of time.
Example 4.1.1: City Growth
Suppose in Flagstaff Arizona, the number of residents increased by 1000 people per year. If the initial population was 46,080 in 1990, can you predict the population in 2013? This is an example of linear growth because the population grows by a constant amount. We list the population in future years below by adding 1000 people for each passing year.
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Year 0 1 2 3 4 5 6 Population 46,080 47,080 48,080 49,080 50,080 51,080 52,
Figure 4.1.1: Graph of Linear Population Growth This is the graph of the population growth over a six year period in Flagstaff, Arizona. It is a straight line and can be modeled with a linear growth model.
The population growth can be modeled with a linear equation. The initial population 𝑃𝑃 0 is 48,080. The future population depends on the number of years, t , after the initial year. The model is P(t) = 46,080 + 1000 t
To predict the population in 2013, we identify how many years it has been from 1990, which is year zero. So n = 23 for the year 2013.
P (23) = 46, 080 +1000(23) =69, 080
The population of Flagstaff in 2013 would be 69,080 people.
Linear Growth Model : Linear growth begins with an initial population called P 0. In each time period or generation t , the population changes by a constant amount called the common difference d. The basic model is: P t ( ) = P 0 + td
43000
44000
45000
46000
47000
48000
49000
50000
51000
52000
53000
1990 1991 1992 1993 1994 1995 1996
Population
Time in Years
City Growth (Linear)
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Example 4.1.3: Car Depreciation
Assume a car depreciates by the same amount each year. Joe purchased a car in 2010 for $16,800. In 2014 it is worth $12,000. Find the linear growth model. Predict how much the car will be worth in 2020.
P 0 (^) = 16,800and P (4) =12, 000 To find the linear growth model for this problem, we need to find the common difference d.
P t ( ) = P 0 + td
12, 000 = 16,800 + 4 d
− 4800 = 4 d
− 1200 = d The common difference of depreciation each year is d = $ − 1200. Thus, the
linear growth model for this problem is: P t ( ) = 16,800 − 1200 t
Now, to find out how much the car will be worth in 2020, we need to know how many years that is from the purchase year. Since it is ten years later, t = 10.
P (10) = 16,800 − 1200(10) = 16,800 −12, 000 =4,
The car is worth $4800 in 2020.
Figure 4.1.3: Graph of Car Value Depreciation
Note: The value of the car over time follows a decreasing straight line.
0
2000
4000
6000
8000
10000
12000
14000
16000
18000
0 2 4 6 8 10 12
Value of Car in Dollars
Time in Years
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Section 4.2: Exponential Growth
The next growth we will examine is exponential growth. Linear growth occur by adding the same amount in each unit of time. Exponential growth happens when an initial population increases by the same percentage or factor over equal time increments or generations. This is known as relative growth and is usually expressed as percentage. For example, let’s say a population is growing by 1.6% each year. For every 1000 people in the population, there will be 1000 × 0.016 = 16 more people added per year.
Exponential Growth: A quantity grows exponentially if it grows by a constant factor or rate for each unit of time.
Figure 4.2.1: Graphical Comparison of Linear and Exponential Growth In this graph, the blue straight line represents linear growth and the red curved line represents exponential growth.
Example 4.2.1: City Growth
A city is growing at a rate of 1.6% per year. The initial population in 2010 is P 0 (^) = 125, 000. Calculate the city’s population over the next few years.
The relative growth rate is 1.6%. This means an additional 1.6% is added on to 100% of the population that already exists each year. This is a factor of 101.6%.
Population in 2011 = 125,000(1.016)^1 = 127,
Population in 2012 = 127,000(1.016) = 125,000(1.016)^2 = 129,
Population in 2013 = 129,032(1.016) = 125,000(1.016)^3 = 131,
0
200
400
600
800
1000
1200
0 1 2 3 4 5 6 7 8 9 10
Population
Time
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64
P = −
Figure 4.2.3: Graph of St. Louis, Missouri Population Decline
Note: The graph of the population of St. Louis, Missouri over time follows a declining exponential growth model.
Exponential Growth Model P t ( ) = P 0 (1 + r ) t 0 is the initial population, is the relative growth rate. is the time unit. is positive if the population is increasing and negative if the population is decreasing
P
r t r
Example 4.2.3: Inflation
The average inflation rate of the U.S. dollar over the last five years is 1.7% per year. If a new car cost $18,000 five years ago, how much would it cost today? (U.S. Inflation Calculator, n.d.) To solve this problem, we use the exponential growth model with r = 1 .7%.
P 0 (^) = 18, 000and t = 5
( ) 18, 000 1( 0.017^ )
P t = + t
0
100000
200000
300000
400000
500000
600000
700000
800000
900000
1950 1960 1970 1980 1990 2000 2010 2020
Population
Time in Years
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P 5 = 18, 000 1 + 0.017 5 =19,582.
This car would cost $19,582.91today.
Example 4.2.4: Ebola Epidemic in Sierra Leone
In May of 2014 there were 15 cases of Ebola in Sierra Leone. By August, there were 850 cases. If the virus is spreading at the same rate (exponential growth), how many cases will there be in February of 2015? (McKenna, 2014)
To solve this problem, we have to find three things; the growth rate per month, the exponential growth model, and the number of cases of Ebola in February 2015. First calculate the growth rate per month. To do this, use the initial population P 0 (^) = 15 , in May 2014. Also, in August, three months later, the number of cases was 850 so, P (3) = 850. Use these values and the exponential growth model to solve for r. 0 3 3 3 3 3
P t P r^ t r r r r r
The growth rate is 284% per month. Thus, the exponential growth model is: P t ( ) = 15(1 + 2.84) t^ =15(3.84) t Now, we use this to calculate the number of cases of Ebola in Sierra Leone in February 2015, which is 9 months after the initial outbreak so, t = 9. P (9) = 15(3.84)^9 =2, 725, 250 If this same exponential growth rate continues, the number of Ebola cases in Sierra Leone in February 2015 would be 2,725,250. This is a bleak prediction for the community of Sierra Leone. Fortunately, the growth rate of this deadly virus should be reduced by the world community and World Health Organization by providing the needed means to fight the initial spread.
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Section 4.3: Special Cases: Doubling Time and Half-Life
Example 4.3.1: April Fool’s Joke
Let’s say that on April 1st^ I say I will give you a penny, on April 2nd^ two pennies, four pennies on April 3rd, and that I will double the amount each day until the end of the month. How much money would I have agreed to give you on April 30? With P 0 (^) = $0.01, we get the following table:
Table 4.3.1: April Fool’s Joke
Day Dollar Amount April 1 = P 0^0. April 2 = P 1^0. April 3 = P 2^0. April 4 = P 3^0. April 5 = P 4^0. April 6 = P 5^0. …. …. April 30 = P 29?
In this example, the money received each day is 100% more than the previous day. If we use the exponential growth model P t ( ) = P 0 (1 + r ) t with r = 1, we get the doubling time model.
0 0
29
We use it to find the dollar amount when 29 which represents April (29) 0.01(2) $5,368, 709.
P t P t^ P t t P
Surprised?
That is a lot of pennies.
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Doubling Time Model:
Example 4.3.2: E. coli Bacteria
A water tank up on the San Francisco Peaks is contaminated with a colony of 80,000 E. coli bacteria. The population doubles every five days. We want to find a model for the population of bacteria present after t days. The amount of time it takes the population to double is five days, so this is our time unit. After t days have passed, then 5 𝑡𝑡 is the number of time units that have passed. Starting with the initial amount of 80,000 bacteria, our doubling model becomes:
( ) 80, 000(2)^5
t P t =
Using this model, how large is the colony in two weeks’ time? We have to be careful that the units on the times are the same; 2 weeks = 14 days. 14 P (14) = 80, 000(2) 5 =557, The colony is now 557,152 bacteria.
Doubling Time Model: If D is the doubling time of a quantity (the amount of time it takes the quantity to double) and P 0 is the initial amount of the quantity then the
amount of the quantity present after t units of time is ( ) 0 (2)
t P t = P D
Example 4.3.3: Flies
The doubling time of a population of flies is eight days. If there are initially 100 flies, how many flies will there be in 17 days? To solve this problem, use the doubling time model with D = 8 and P 0 (^) = 100 so the doubling time model for this problem is:
( ) 100(2)^8
t P t = When t = 17 days, 17 P (17) = 100(2) 8 = 436
There are 436 flies after 17 days.
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log 6.25 log 2 6
^ t =
Now, calculate log 6.25 and log 2 with your calculator.
^ t = (^) ×
0.3010 6
t
6
=^ t
t = 15. The population would increase to 25,000 bacteria in approximately 15.9 hours.
Rule of 70:
There is a simple formula for approximating the doubling time of a population. It is called the rule of 70 and it is an approximation for growth rates less than 15%. Do not use this formula if the growth rate is 15% or greater.
Rule of 70 : For a quantity growing at a constant percentage rate (not written as a decimal), R , per time period, the doubling time is approximately given by 70 Doubling time D R
Example 4.3.5: Bird Population
A bird population on a certain island has an annual growth rate of 2.5% per year. Approximate the number of years it will take the population to double. If the initial population is 20 birds, use it to find the bird population of the island in 17 years.
To solve this problem, first approximate the population doubling time. 70 Doubling time 28
D ≈ = years.
With the bird population doubling in 28 years, we use the doubling time model to find the population is 17 years. 28
17 28
When 17 years
(16) 20(2) 30.
t P t t
P
There will be 30 birds on the island in 17 years.
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Example 4.3.6: Cancer Growth Rate
A certain cancerous tumor doubles in size every six months. If the initial size of the tumor is four cells, how many cells will there in three years? In seven years?
To calculate the number of cells in the tumor, we use the doubling time model. Change the time units to be the same. The doubling time is six months = 0. years.
3
When 3 years
(3) 4(2) 256 cells
t P t t
P
7
When 7 years
(7) 4(2) 65,536 cells
t
P
Example 4.3.7: Approximating Annual Growth Rate
Suppose that a certain city’s population doubles every 12 years. What is the approximate annual growth rate of the city?
By solving the doubling time model for the growth rate, we can solve this problem. 70 D R
R D R
R
RD ≈ 70
RD 70
D D
Annual growth rate (^70) 5.83% 12
R
D
R
The annual growth rate of the city is approximately 5.83%
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Example 4.3.9: Lead-
Lead-209 is a radioactive isotope. It has a half-life of 3.3 hours. Suppose that 40 milligrams of this isotope is created in an experiment, how much is left after 14 hours? Use the half-life model to solve this problem.
P 0 (^) = 40 and H = 3.3, so the half-life model for this problem is:
t P t = ^
when t = 14 hours,
14 1 3. 14 40 2. 2
P = ^ =
There are 2.1 milligrams of Lead-209 remaining after 14 hours.
Figure 4.3.3: Lead-209 Decay Graph
Note: The milligrams of Lead-209 remaining follows a decreasing exponential growth model.
Example 4.3.10: Nobelium-
Nobelium-259 has a half-life of 58 minutes. If you have 1000 grams, how much will be left in two hours? We solve this problem using the half-life model. Before we begin, it is important to note the time units. The half-life is given in minutes and we want to know how much is left in two hours. Convert hours to minutes when using the model: two hours = 120 minutes.
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0 2 4 6 8 10 12 14 16
Milligrams of Lead-
Time in Hours
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P 0 (^) = 1000 and H = 58 minutes, so the half-life model for this problem is:
t P t = ^ When t = 120 minutes,
120 1 58 120 1000 238. 2
P = ^ =
There are 238 grams of Nobelium-259 is remaining after two hours.
Example 4.3.11: Carbon- Radioactive carbon-14 is used to determine the age of artifacts because it concentrates in organisms only when they are alive. It has a half-life of 5730 years. In 1947, earthenware jars containing what are known as the Dead Sea Scrolls were found. Analysis indicated that the scroll wrappings contained 76% of their original carbon-14. Estimate the age of the Dead Sea Scrolls. In this problem, we want to estimate the age of the scrolls. In 1947, 76% of the carbon-14 remained. This means that the amount remaining at time t divided by the original amount of carbon-14, P 0 , is equal to 76%. So, 0
P t P
= we use this fact to solve for t.
5730 0
t P t = P ^
0
t P t P
=^ ^
t =^ ^
Now, take the log of both sides of the equation.
log 0.76 log 2
t = ^ The exponent comes down using rules of logarithms.
1 log 0.76 log 5730 2
^ t =
Now, calculate (^) log 0.76 and log^1 2
with your calculator.
− = ^^ t × −
0.3010 5730
− (^) = t −
5730
=^ t
t = 2269.08. The Dead Sea Scrolls are well over 2000 years old.
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Rule of 70 for Half-Life:
There is simple formula for approximating the half-life of a population. It is called the rule of 70 and is an approximation for decay rates less than 15%. Do not use this formula if the decay rate is 15% or greater.
Rule of 70 : For a quantity decreasing at a constant percentage (not written as a decimal), R , per time period, the half-life is approximately given by: Half-life H^70 R
Example 4.3.13: Elephant Population
The population of wild elephants is decreasing by 7% per year. Approximate the half -life for this population. If there are currently 8000 elephants left in the wild, how many will remain in 25 years? To solve this problem, use the half-life approximation formula.
0 10
25 10
Half-life H 70 10 years 7 7000, 10 years, so the half-life model for this problem is:
When 25,
(25) 7000 1 1237. 2
t
P H
P t
t
P
= ^
= ^ =
There will be approximately 1237 wild elephants left in 25 years.
Figure 4.3.4: Elephant Population over a 70 Year Span.
0
1000
2000
3000
4000
5000
6000
7000
8000
0 20 40 60 80
Elephant Population
Time in Years
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Note: The population of elephants follows a decreasing exponential growth model.
Review of Exponent Rules and Logarithm Rules
Rules of Exponents Rules of Logarithm for the Common Logarithm (Base 10) Definition of an Exponent ..... ( 's multiplied together)
a^ n a a a a a n a
Definition of a Logarithm 10 y^ = x if and only if log x = y
Zero Rule a^0 = 1
Product Rule am ⋅ an = am + n Product Rule^ log ( xy )= log x +log y
Quotient Rule
m m n n
a (^) a a
= − Quotient Rule log^ x^ log x log y y
=^ −
Power Rule ( )
n m n m a = a ⋅ Power Rule^ log^ log^ (^ 0) xr = r x x > Distributive Rules
n (^) n n (^) n n n
a a ab a b b b
= ^ =
log10 x^ = x log10= x
Negative Exponent Rules 1 ,
n n n n
a b a a b a
− − (^) = ^ ^ =^
10 log x^ = x ( x >0)
Section 4.4: Natural Growth and Logistic Growth
In this chapter, we have been looking at linear and exponential growth. Another very useful tool for modeling population growth is the natural growth model. This model uses base e , an irrational number, as the base of the exponent instead of (^) (1 + r ). You may remember
learning about e in a previous class, as an exponential function and the base of the natural logarithm.
The Natural Growth Model: (^) P t ( ) = P e 0 kt where P 0 is the initial population, k is the growth rate per unit of time, and t is the number of time periods.
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