Examples

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?

Answer: 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. [latex-display]d=slope=\frac{\text{changeinoutput}}{\text{changeininput}}=\frac{15,000-12,000}{2007-2003}=\frac{3000}{4}=750[/latex-display] We can now write our equation in whichever form is preferred.

Recursive form

P­₀ = 12,000

P_­n = P_­n-1 + 750

Explicit form

P­_n = 12,000 + 750_n

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

View more about this example here. https://youtu.be/J1XqqlKzYGs

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Gasoline consumption in the US has been increasing steadily. Consumption data from 1992 to 2004 is shown below.[footnote]http://www.bts.gov/publications/national_transportation_statistics/2005/html/table_04_10.html[/footnote] 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?

Year	'92	'93	'94	'95	'96	'97	'98	'99	'00	'01	'02	'03	'04
Consumption (billion of gallons)	110	111	113	116	118	119	123	125	126	128	131	133	136

Answer: 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. Equivalently, [latex]d=slope=\frac{\text{changeinoutput}}{\text{changeininput}}=\frac{133-111}{10-0}=\frac{22}{10}=2.2[/latex]billion gallons per year We can now write our equation in whichever form is preferred.

Recursive form

P­₀ = 111

P­_n = P­_n-1 + 2.2

Explicit form

P­_n = 111 + 2.2_n

  Calculating values using the explicit form and plotting them with the original data shows how well our model fits the data. 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)

P_­23 = 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 P­_n = 200, and solve for n:

P_n = 200                                  Replace P_n with our model

111 + 2.2_n = 200                    Subtract 111 from both sides

2.2_n = 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 steps for reaching this answer are detailed in the following video. https://youtu.be/ApFxDWd6IbE

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The cost, in dollars, of a gym membership for n months can be described by the explicit equation P­_n = 70 + 30_n. What does this equation tell us?

Answer: 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.

The explanation for this example is detailed below. https://youtu.be/0Uwz5dmLTtk

Examples

A friend is using the equation P­_n = 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?

Answer: In the equation, P_­0 = 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.

View the following to see this example worked out. https://youtu.be/T8Yz94De5UM

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In 1990, 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[footnote]http://www.eia.doe.gov/oiaf/1605/ggrpt/carbon.html[/footnote]. If the emissions grow exponentially and continue at the same rate, what will the emissions grow to by 2050?

Answer: 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 P_­60

P_­60 = (1+0.0208)⁶⁰ 962 = 3308.4 million metric tons of CO₂ in 2050

View more about this example here. https://youtu.be/9Zu2uONfLkQ

Example

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?

Answer: Again we will get n = 0 correspond with 1990, giving P_­0 = 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 = P_­0 + 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.

A demonstration of this example can be seen in the following video. https://youtu.be/yiuZoiRMtYM