66 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

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.

Example 5

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 2014?

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 people.

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?

Example 6

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.

Example 7

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 tons^[1]. 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.

[latex]\frac{1182}{962}={{(1+r)}^{10}}[/latex]                             Take the 10th root of both sides

[latex]\sqrt[10]{\frac{1182}{962}}=1+r[/latex]                                Subtract 1 from both sides

[latex]r=\sqrt[10]{\frac{1182}{962}}-1=0.0208[/latex] = 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

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 then expect two years after they went public?

Example 8

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

We know the value for P­0, so we can put that into the equation:

P­10 = 962 + 10d

Since we know that P­10 = 1182, substituting that in we get

1182 = 962 + 10d

We can now solve this equation for the common difference, d.

1182 – 962 = 10d

220 = 10d

d = 22

This tells us that if the emissions are changing linearly, they are growing by 22 million metric tons each year. Predicting the emissions in 2050,

P­60 = 962 + 22(60) = 2282 million metric tons.

You will notice that this number is substantially smaller than the prediction from the exponential growth model. Calculating and plotting more values helps illustrate the differences.

So how do we know which growth model to use when working with data? There are two approaches which should be used together whenever possible:

• 1) Find more than two pieces of data. Plot the values, and look for a trend. Does the data appear to be changing like a line, or do the values appear to be curving upwards?
• 2) Consider the factors contributing to the data. Are they things you would expect to change linearly or exponentially? For example, in the case of carbon emissions, we could expect that, absent other factors, they would be tied closely to population values, which tend to change exponentially.