31 Oct Growth Models 173 David Lippman Creative Commons BY-SA Growth Models Populations of people, animals, and items are gro
Growth Models exercises 1, 3, 5, 7, 9, 11, and 14.
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Growth Models 173
© David Lippman Creative Commons BY-SA
Growth Models Populations of people, animals, and items are growing all around us. By understanding how
things grow, we can better understand what to expect in the future. In this chapter, we focus
on time-dependant change.
Linear (Algebraic) Growth Marco is a collector of antique soda bottles. His collection currently contains 437 bottles.
Every year, he budgets enough money to buy 32 new bottles. Can we determine how many
bottles he will have in 5 years, and how long it will take for his collection to reach 1000
While both of these questions you could probably solve without an equation or formal
mathematics, we are going to formalize our approach to this problem to provide a means to
answer more complicated questions.
Suppose that Pn represents the number, or population, of bottles Marco has after n years. So
P0 would represent the number of bottles now, P1 would represent the number of bottles after
1 year, P2 would represent the number of bottles after 2 years, and so on. We could describe
how Marco’s bottle collection is changing using:
P0 = 437
Pn = Pn-1 + 32
This is called a recursive relationship. A recursive relationship is a formula which relates
the next value in a sequence to the previous values. Here, the number of bottles in year n can
be found by adding 32 to the number of bottles in the previous year, Pn-1. Using this
relationship, we could calculate:
P1 = P0 + 32 = 437 + 32 = 469
P2 = P1 + 32 = 469 + 32 = 501
P3 = P2 + 32 = 501 + 32 = 533
P4 = P3 + 32 = 533 + 32 = 565
P5 = P4 + 32 = 565 + 32 = 597
We have answered the question of how many bottles Marco will have in 5 years. However,
solving how long it will take for his collection to reach 1000 bottles would require a lot more
While recursive relationships are excellent for describing simply and cleanly how a quantity
is changing, they are not convenient for making predictions or solving problems that stretch
far into the future. For that, a closed or explicit form for the relationship is preferred. An
explicit equation allows us to calculate Pn directly, without needing to know Pn-1. While
you may already be able to guess the explicit equation, let us derive it from the recursive
formula. We can do so by selectively not simplifying as we go:
P1 = 437 + 32 = 437 + 1(32)
P2 = P1 + 32 = 437 + 32 + 32 = 437 + 2(32)
P3 = P2 + 32 = (437 + 2(32)) + 32 = 437 + 3(32)
P4 = P3 + 32 = (437 + 3(32)) + 32 = 437 + 4(32)
You can probably see the pattern now, and generalize that
Pn = 437 + n(32) = 437 + 32n
Using this equation, we can calculate how many bottles he’ll have after 5 years:
P5 = 437 + 32(5) = 437 + 160 = 597
We can now also solve for when the collection will reach 1000 bottles by substituting in
1000 for Pn and solving for n
1000 = 437 + 32n
563 = 32n
n = 563/32 = 17.59
So Marco will reach 1000 bottles in 18 years.
In the previous example, Marco’s collection grew by
the same number of bottles every year. This constant
change is the defining characteristic of linear growth.
Plotting the values we calculated for Marco’s
collection, we can see the values form a straight line,
the shape of linear growth.
If a quantity starts at size P0 and grows by d every time period, then the quantity after n
time periods can be determined using either of these relations:
Pn = Pn-1 + d
Pn = P0 + d n
In this equation, d represents the common difference – the amount that the population
changes each time n increases by 1
Connection to prior learning – slope and intercept
You may recognize the common difference, d, in our linear equation as slope. In fact, the
entire explicit equation should look familiar – it is the same linear equation you learned in
algebra, probably stated as y = mx + b.
0 1 2 3 4 5
Years from now
Growth Models 175
In the standard algebraic equation y = mx + b, b was the y-intercept, or the y value when x
was zero. In the form of the equation we’re using, we are using P0 to represent that initial
In the y = mx + b equation, recall that m was the slope. You might remember this as “rise
over run”, or the change in y divided by the change in x. Either way, it represents the same
thing as the common difference, d, we are using – the amount the output Pn changes when
the input n increases by 1.
The equations y = mx + b and Pn = P0 + d n mean the same thing and can be used the same
ways, we’re just writing it somewhat differently.
The population of elk in a national forest was measured to be 12,000 in 2003, and was
measured again to be 15,000 in 2007. If the population continues to grow linearly at this
rate, what will the elk population be in 2014?
To begin, we need to define how we’re going to measure n. Remember that P0 is the
population when n = 0, so we probably don’t want to literally use the year 0. Since we
already know the population in 2003, let us define n = 0 to be the year 2003. Then
P0 = 12,000.
Next we need to find d. Remember d is the growth per time period, in this case growth per
year. Between the two measurements, the population grew by 15,000-12,000 = 3,000, but it
took 2007-2003 = 4 years to grow that much. To find the growth per year, we can divide:
3000 elk / 4 years = 750 elk in 1 year.
Alternatively, you can use the slope formula from algebra to determine the common
difference, noting that the population is the output of the formula, and time is the input.
change in output 15,000 12,000 3000 slope 750
change in input 2007 2003 4 d
− = = = = =
We can now write our equation in whichever form is preferred.
P0 = 12,000
Pn = Pn-1 + 750
Pn = 12,000 + 750n
To answer the question, we need to first note that the year 2014 will be n = 11, since 2014 is
11 years after 2003. The explicit form will be easier to use for this calculation:
P11 = 12,000 + 750(11) = 20,250 elk
Gasoline consumption in the US has been increasing steadily. Consumption data from 1992
to 2004 is shown below1. Find a model for this data, and use it to predict consumption in
2016. If the trend continues, when will consumption reach 200 billion gallons?
Plotting this data, it appears to have an
approximately linear relationship:
While there are more advanced statistical
techniques that can be used to find an equation
to model the data, to get an idea of what is
happening, we can find an equation by using
two pieces of the data – perhaps the data from
1993 and 2003.
Letting n = 0 correspond with 1993 would
give P0 = 111 billion gallons.
To find d, we need to know how much the gas consumption increased each year, on average.
From 1993 to 2003 the gas consumption increased from 111 billion gallons to 133 billion
gallons, a total change of 133 – 111 = 22 billion gallons, over 10 years. This gives us an
average change of 22 billion gallons / 10 year = 2.2 billion gallons per year.
change in output 133 111 22 2.2
change in input 10 0 10 d slope
− = = = = =
− billion gallons per year
We can now write our equation in whichever form is preferred.
P0 = 111
Pn = Pn-1 + 2.2
Pn = 111 + 2.2n
Calculating values using the explicit form and
plotting them with the original data shows how
well our model fits the data.
Year ‘92 ‘93 ‘94 ‘95 ‘96 ‘97 ‘98 ‘99 ‘00 ‘01 ‘02 ‘03 ‘04
gallons) 110 111 113 116 118 119 123 125 126 128 131 133 136
1992 1996 2000 2004
G a s C
1992 1996 2000 2004
Growth Models 177
We can now use our model to make predictions about the future, assuming that the previous
trend continues unchanged. To predict the gasoline consumption in 2016:
n = 23 (2016 – 1993 = 23 years later)
P23 = 111 + 2.2(23) = 161.6
Our model predicts that the US will consume 161.6 billion gallons of gasoline in 2016 if the
current trend continues.
To find when the consumption will reach 200 billion gallons, we would set Pn = 200, and
solve for n:
Pn = 200 Replace Pn with our model
111 + 2.2n = 200 Subtract 111 from both sides
2.2n = 89 Divide both sides by 2.2
n = 40.4545
This tells us that consumption will reach 200 billion about 40 years after 1993, which would
be in the year 2033.
The cost, in dollars, of a gym membership for n months can be described by the explicit
equation Pn = 70 + 30n. What does this equation tell us?
The value for P0 in this equation is 70, so the initial starting cost is $70. This tells us that
there must be an initiation or start-up fee of $70 to join the gym.
The value for d in the equation is 30, so the cost increases by $30 each month. This tells us
that the monthly membership fee for the gym is $30 a month.
Try it Now 1
The number of stay-at-home fathers in Canada has been growing steadily2. While the trend
is not perfectly linear, it is fairly linear. Use the data from 1976 and 2010 to find an explicit
formula for the number of stay-at-home fathers, then use it to predict the number if 2020.
When good models go bad
When using mathematical models to predict future behavior, it is important to keep in mind
that very few trends will continue indefinitely.
Year 1976 1984 1991 2000 2010
Number of stay-at-home fathers 20,610 28,725 43,530 47,665 53,555
Suppose a four year old boy is currently 39 inches tall, and you are told to expect him to
grow 2.5 inches a year.
We can set up a growth model, with n = 0 corresponding to 4 years old.
P0 = 39
Pn = Pn-1 + 2.5
Pn = 39 + 2.5n
So at 6 years old, we would expect him to be
P2 = 39 + 2.5(2) = 44 inches tall
Any mathematical model will break down eventually. Certainly, we shouldn’t expect this
boy to continue to grow at the same rate all his life. If he did, at age 50 he would be
P46 = 39 + 2.5(46) = 154 inches tall = 12.8 feet tall!
When using any mathematical model, we have to consider which inputs are reasonable to
use. Whenever we extrapolate, or make predictions into the future, we are assuming the
model will continue to be valid.
Exponential (Geometric) Growth Suppose that every year, only 10% of the fish in a lake have surviving offspring. If there
were 100 fish in the lake last year, there would now be 110 fish. If there were 1000 fish in
the lake last year, there would now be 1100 fish. Absent any inhibiting factors, populations
of people and animals tend to grow by a percent of the existing population each year.
Suppose our lake began with 1000 fish, and 10% of the fish have surviving offspring each
year. Since we start with 1000 fish, P0 = 1000. How do we calculate P1? The new
population will be the old population, plus an additional 10%. Symbolically:
P1 = P0 + 0.10P0
Notice this could be condensed to a shorter form by factoring:
P1 = P0 + 0.10P0 = 1P0 + 0.10P0 = (1+ 0.10)P0 = 1.10P0
While 10% is the growth rate, 1.10 is the growth multiplier. Notice that 1.10 can be
thought of as “the original 100% plus an additional 10%”
For our fish population,
P1 = 1.10(1000) = 1100
Growth Models 179
We could then calculate the population in later years:
P2 = 1.10P1 = 1.10(1100) = 1210
P3 = 1.10P2 = 1.10(1210) = 1331
Notice that in the first year, the population grew by 100 fish, in the second year, the
population grew by 110 fish, and in the third year the population grew by 121 fish.
While there is a constant percentage growth, the actual increase in number of fish is
increasing each year.
Graphing these values we see that this growth
doesn’t quite appear linear.
To get a better picture of how this percentage-
based growth affects things, we need an explicit
form, so we can quickly calculate values further
out in the future.
Like we did for the linear model, we will start
building from the recursive equation:
P1 = 1.10P0 = 1.10(1000)
P2 = 1.10P1 = 1.10(1.10(1000)) = 1.102(1000)
P3 = 1.10P2 = 1.10(1.102(1000)) = 1.103(1000)
P4 = 1.10P3 = 1.10(1.103(1000)) = 1.104(1000)
Observing a pattern, we can generalize the explicit form to be:
Pn = 1.10n(1000), or equivalently, Pn = 1000(1.10n)
From this, we can quickly calculate the number of
fish in 10, 20, or 30 years:
P10 = 1.1010(1000) = 2594
P20 = 1.1020(1000) = 6727
P30 = 1.1030(1000) = 17449
Adding these values to our graph reveals a shape
that is definitely not linear. If our fish population
had been growing linearly, by 100 fish each year,
the population would have only reached 4000 in 30
years compared to almost 18000 with this percent-
based growth, called exponential growth.
In exponential growth, the population grows proportional to the size of the population, so as
the population gets larger, the same percent growth will yield a larger numeric growth.
0 1 2 3 4 5
Years from now
0 5 10 15 20 25 30
Years from now
If a quantity starts at size P0 and grows by R% (written as a decimal, r) every time
period, then the quantity after n time periods can be determined using either of these
Pn = (1+r) Pn-1
Pn = (1+r)n P0 or equivalently, Pn = P0 (1+r)n
We call r the growth rate.
The term (1+r) is called the growth multiplier, or common ratio.
Between 2007 and 2008, Olympia, WA grew almost 3% to a population of 245 thousand
people. If this growth rate was to continue, what would the population of Olympia be in
As we did before, we first need to define what year will correspond to n = 0. Since we know
the population in 2008, it would make sense to have 2008 correspond to n = 0, so P0 =
245,000. The year 2014 would then be n = 6.
We know the growth rate is 3%, giving r = 0.03.
Using the explicit form:
P6 = (1+0.03)6 (245,000) = 1.19405(245,000) = 292,542.25
The model predicts that in 2014, Olympia would have a population of about 293 thousand
Evaluating exponents on the calculator
To evaluate expressions like (1.03)6, it will be easier to use a calculator than multiply
1.03 by itself six times. Most scientific calculators have a button for exponents. It is
typically either labeled like:
^ , yx , or xy .
To evaluate 1.036 we’d type 1.03 ^ 6, or 1.03 yx 6. Try it out – you should get an
answer around 1.1940523.
Growth Models 181
Try it Now 2
India is the second most populous country in the world, with a population in 2008 of about
1.14 billion people. The population is growing by about 1.34% each year. If this trend
continues, what will India’s population grow to by 2020?
A friend is using the equation Pn = 4600(1.072)n to predict the annual tuition at a local
college. She says the formula is based on years after 2010. What does this equation tell us?
In the equation, P0 = 4600, which is the starting value of the tuition when n = 0. This tells us
that the tuition in 2010 was $4,600.
The growth multiplier is 1.072, so the growth rate is 0.072, or 7.2%. This tells us that the
tuition is expected to grow by 7.2% each year.
Putting this together, we could say that the tuition in 2010 was $4,600, and is expected to
grow by 7.2% each year.
In 1990, the residential energy use in the US was responsible for 962 million metric tons of
carbon dioxide emissions. By the year 2000, that number had risen to 1182 million metric
tons3. If the emissions grow exponentially and continue at the same rate, what will the
emissions grow to by 2050?
Similar to before, we will correspond n = 0 with 1990, as that is the year for the first piece of
data we have. That will make P0 = 962 (million metric tons of CO2). In this problem, we are
not given the growth rate, but instead are given that P10 = 1182.
When n = 10, the explicit equation looks like:
P10 = (1+r)10 P0
We know the value for P0, so we can put that into the equation:
P10 = (1+r)10 962
We also know that P10 = 1182, so substituting that in, we get
1182 = (1+r)10 962
We can now solve this equation for the growth rate, r. Start by dividing by 962.
101182 (1 )
962 r= + Take the 10th root of both sides
r= + Subtract 1 from both sides
1 0.0208 962
r = − = = 2.08%
So if the emissions are growing exponentially, they are growing by about 2.08% per year.
We can now predict the emissions in 2050 by finding P60
P60 = (1+0.0208)60 962 = 3308.4 million metric tons of CO2 in 2050
As a note on rounding, notice that if we had rounded the growth rate to 2.1%, our
calculation for the emissions in 2050 would have been 3347. Rounding to 2% would
have changed our result to 3156. A very small difference in the growth rates gets
magnified greatly in exponential growth. For this reason, it is recommended to round
the growth rate as little as possible.
If you need to round, keep at least three significant digits – numbers after any leading
zeros. So 0.4162 could be reasonably rounded to 0.416. A growth rate of 0.001027
could be reasonably rounded to 0.00103.
Evaluating roots on the calculator
In the previous example, we had to calculate the 10th root of a number. This is
different than taking the basic square root, √. Many scientific calculators have a button
for general roots. It is typically labeled like:
n , x , or y
To evaluate the 3rd root of 8, for example, we’d either type 3 x 8, or 8 x 3,
depending on the calculator. Try it on yours to see which to use – you should get an
answer of 2.
If your calculator does not have a general root button, all is not lost. You can instead
use the property of exponents which states that 1/nn a a= . So, to compute the 3rd root
of 8, you could use your calculator’s exponent key to evaluate 1/38 . To do this, type:
8 yx ( 1 ÷ 3 )
The parentheses tell the calculator to divide 1/3 before doing the exponent.
Try it Now 3
The number of users on a social networking site was 45 thousand in February when they
officially went public, and grew to 60 thousand by October. If the site is growing
exponentially, and growth continues at the same rate, how many users should they expect
two years after they went public?
Growth Models 183
Looking back at the last example, for the sake of comparison, what would the carbon
emissions be in 2050 if emissions grow linearly at the same rate?
Again we will get n = 0 correspond with 1990, giving P0 = 962. To find d, we could take the
same approach as earlier, noting that the emissions increased by 220 million metric tons in 10
years, giving a common difference of 22 million metric tons each year.
Alternatively, we could use an approach similar to that which we used to find the exponential
equation. When n = 10, the explicit linear equation looks like:
P10 = P0 + 10d
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