Example: Using a Linear Model to Investigate a Town’s Population

A town’s population has been growing linearly. In 2004 the population was 6,200. By 2009 the population had grown to 8,100. Assume this trend continues.

1. Predict the population in 2013.
2. Identify the year in which the population will reach 15,000.

Answer: The two changing quantities are the population size and time. While we could use the actual year value as the input quantity, doing so tends to lead to very cumbersome equations because the y-intercept would correspond to the year 0, more than 2000 years ago! To make computation a little nicer, we will define our input as the number of years since 2004:

• Input: t, years since 2004
• Output: P(t), the town’s population

To predict the population in 2013 (t = 9), we would first need an equation for the population. Likewise, to find when the population would reach 15,000, we would need to solve for the input that would provide an output of 15,000. To write an equation, we need the initial value and the rate of change, or slope. To determine the rate of change, we will use the change in output per change in input.

[latex]m=\frac{\text{change in output}}{\text{change in input}}[/latex]

The problem gives us two input-output pairs. Converting them to match our defined variables, the year 2004 would correspond to [latex]t=0[/latex], giving the point [latex]\left(0,\text{6200}\right)[/latex]. Notice that through our clever choice of variable definition, we have "given" ourselves the y-intercept of the function. The year 2009 would correspond to [latex]t=\text{5}[/latex], giving the point [latex]\left(5,\text{8100}\right)[/latex]. The two coordinate pairs are [latex]\left(0,\text{6200}\right)[/latex] and [latex]\left(5,\text{8100}\right)[/latex]. Recall that we encountered examples in which we were provided two points earlier in the chapter. We can use these values to calculate the slope.

[latex]\begin{array}{l} m=\frac{8100 - 6200}{5 - 0}\hfill \\ \text{ }=\frac{1900}{5}\hfill \\ \text{ }=380\text{ people per year}\hfill \end{array}[/latex]

We already know the y-intercept of the line, so we can immediately write the equation: [latex-display]P\left(t\right)=380t+6200[/latex-display] To predict the population in 2013, we evaluate our function at t = 9.

[latex]\begin{array}{l}P\left(9\right)=380\left(9\right)+6,200\hfill \\ \text{ }=9,620\hfill \end{array}[/latex]

If the trend continues, our model predicts a population of 9,620 in 2013. To find when the population will reach 15,000, we can set [latex]P\left(t\right)=15000[/latex] and solve for t.

[latex]\begin{array}{l}15000=380t+6200\hfill \\ \text{ }8800=380t\hfill \\ \text{ }t\approx 23.158\hfill \end{array}[/latex]

Our model predicts the population will reach 15,000 in a little more than 23 years after 2004, or somewhere around the year 2027.

Example: Using a Diagram to Model Distance Walked

Anna and Emanuel start at the same intersection. Anna walks east at 4 miles per hour while Emanuel walks south at 3 miles per hour. They are communicating with a two-way radio that has a range of 2 miles. How long after they start walking will they fall out of radio contact?

Answer: In essence, we can partially answer this question by saying they will fall out of radio contact when they are 2 miles apart, which leads us to ask a new question: "How long will it take them to be 2 miles apart?" In this problem, our changing quantities are time and position, but ultimately we need to know how long will it take for them to be 2 miles apart. We can see that time will be our input variable, so we’ll define our input and output variables.

• Input: t, time in hours.
• Output: [latex]A\left(t\right)[/latex], distance in miles, and [latex]E\left(t\right)[/latex], distance in miles

Because it is not obvious how to define our output variable, we’ll start by drawing a picture. Initial Value: They both start at the same intersection so when [latex]t=0[/latex], the distance traveled by each person should also be 0. Thus the initial value for each is 0. Rate of Change: Anna is walking 4 miles per hour and Emanuel is walking 3 miles per hour, which are both rates of change. The slope for A is 4 and the slope for E is 3. Using those values, we can write formulas for the distance each person has walked.

[latex]\begin{array}{l}A\left(t\right)=4t\\ E\left(t\right)=3t\end{array}[/latex]

For this problem, the distances from the starting point are important. To notate these, we can define a coordinate system, identifying the "starting point" at the intersection where they both started. Then we can use the variable, A, which we introduced above, to represent Anna’s position, and define it to be a measurement from the starting point in the eastward direction. Likewise, can use the variable, E, to represent Emanuel’s position, measured from the starting point in the southward direction. Note that in defining the coordinate system, we specified both the starting point of the measurement and the direction of measure. We can then define a third variable, D, to be the measurement of the distance between Anna and Emanuel. Showing the variables on the diagram is often helpful. Recall that we need to know how long it takes for D, the distance between them, to equal 2 miles. Notice that for any given input t, the outputs A(t), E(t), and D(t) represent distances. This picture shows us that we can use the Pythagorean Theorem because we have drawn a right angle. Using the Pythagorean Theorem, we get:

[latex]\begin{array}{l}D{\left(t\right)}^{2}=A{\left(t\right)}^{2}+E{\left(t\right)}^{2}\hfill & \hfill \\ ={\left(4t\right)}^{2}+{\left(3t\right)}^{2}\hfill & \hfill \\ =16{t}^{2}+9{t}^{2}\hfill & \hfill \\ =25{t}^{2}\hfill & \hfill \\ \text{ }D\left(t\right)=\pm \sqrt{25{t}^{2}}\hfill & \text{Solve for }D\left(t\right)\text{ using the square root}\hfill \\ =\pm 5|t|\hfill & \hfill \end{array}[/latex]

In this scenario we are considering only positive values of [latex]t[/latex], so our distance D(t) will always be positive. We can simplify this answer to D(t) = 5t. This means that the distance between Anna and Emanuel is also a linear function. Because D is a linear function, we can now answer the question of when the distance between them will reach 2 miles. We will set the output D(t) = 2 and solve for t.

[latex]\begin{array}{l}D\left(t\right)=2\hfill \\ \text{ }5t=2\hfill \\ \text{ }t=\frac{2}{5}=0.4\hfill \end{array}[/latex]

They will fall out of radio contact in 0.4 hours, or 24 minutes.

Example: Using a Diagram to Model Distance between Cities

There is a straight road leading from the town of Westborough to Agritown 30 miles east and 10 miles north. Partway down this road, it junctions with a second road, perpendicular to the first, leading to the town of Eastborough. If the town of Eastborough is located 20 miles directly east of the town of Westborough, how far is the road junction from Westborough?

Answer: It might help here to draw a picture of the situation. It would then be helpful to introduce a coordinate system. While we could place the origin anywhere, placing it at Westborough seems convenient. This puts Agritown at coordinates (30, 10), and Eastborough at (20, 0). Using this point along with the origin, we can find the slope of the line from Westborough to Agritown:

[latex]m=\frac{10 - 0}{30 - 0}=\frac{1}{3}[/latex]

The equation of the road from Westborough to Agritown would be

[latex]W\left(x\right)=\frac{1}{3}x[/latex]

From this, we can determine the perpendicular road to Eastborough will have slope [latex]m=-3[/latex]. Because the town of Eastborough is at the point (20, 0), we can find the equation:

[latex]\begin{array}{l}E\left(x\right)=-3x+b\hfill & \hfill \\ 0=-3\left(20\right)+b\hfill & \text{Substitute in (20, 0)}\hfill \\ b=60\hfill & \hfill \\ E\left(x\right)=-3x+60\hfill & \hfill \end{array}[/latex]

We can now find the coordinates of the junction of the roads by finding the intersection of these lines. Setting them equal,

[latex]\begin{array}{l}\text{ }\frac{1}{3}x=-3x+60\hfill & \hfill \\ \frac{10}{3}x=60\hfill & \hfill \\ 10x=180\hfill & \hfill \\ \text{ }x=18\hfill & \text{Substituting this back into }W\left(x\right)\hfill \\ \text{ }y=W\left(18\right)\hfill & \hfill \\ \text{ }=\frac{1}{3}\left(18\right)\hfill & \hfill \\ \text{ }=6\hfill & \hfill \end{array}[/latex]

The roads intersect at the point (18, 6). Using the distance formula, we can now find the distance from Westborough to the junction.

[latex]\begin{array}{l}\text{distance}=\sqrt{{\left({x}_{2}-{x}_{1}\right)}^{2}+{\left({y}_{2}-{y}_{1}\right)}^{2}}\hfill \\ \text{ }=\sqrt{{\left(18 - 0\right)}^{2}+{\left(6 - 0\right)}^{2}}\hfill \\ \text{ }\approx 18.974\text{ miles}\hfill \end{array}[/latex]

Analysis of the Solution

One nice use of linear models is to take advantage of the fact that the graphs of these functions are lines. This means real-world applications discussing maps need linear functions to model the distances between reference points.

 Example: Using a Scatter Plot to Investigate Cricket Chirps

The table below shows the number of cricket chirps in 15 seconds, for several different air temperatures, in degrees Fahrenheit.[footnote]Selected data from http://classic.globe.gov/fsl/scientistsblog/2007/10/. Retrieved Aug 3, 2010[/footnote] Plot this data, and determine whether the data appears to be linearly related.

Chirps	44	35	20.4	33	31	35	18.5	37	26
Temperature	80.5	70.5	57	66	68	72	52	73.5	53

Answer: Plotting this data suggests that there may be a trend. We can see from the trend in the data that the number of chirps increases as the temperature increases. The trend appears to be roughly linear, though certainly not perfectly so.

Example: Finding a Line of Best Fit

Find a linear function that fits the data in the table below by "eyeballing" a line that seems to fit.

Chirps	44	35	20.4	33	31	35	18.5	37	26
Temperature	80.5	70.5	57	66	68	72	52	73.5	53

Answer: On a graph, we could try sketching a line. Using the starting and ending points of our hand drawn line, points (0, 30) and (50, 90), this graph has a slope of [latex]m=\frac{60}{50}=1.2[/latex] and a y-intercept at 30. This gives an equation of [latex]T\left(c\right)=1.2c+30[/latex] where c is the number of chirps in 15 seconds, and T(c) is the temperature in degrees Fahrenheit. The resulting equation is represented in the graph below.

Analysis of the Solution

This linear equation can then be used to approximate answers to various questions we might ask about the trend.

Example: Understanding Interpolation and Extrapolation

Chirps	44	35	20.4	33	31	35	18.5	37	26
Temperature	80.5	70.5	57	66	68	72	52	73.5	53

Use the cricket data above to answer the following questions:

1. Would predicting the temperature when crickets are chirping 30 times in 15 seconds be interpolation or extrapolation? Make the prediction, and discuss whether it is reasonable.
2. Would predicting the number of chirps crickets will make at 40 degrees be interpolation or extrapolation? Make the prediction, and discuss whether it is reasonable.

Answer:

1. The number of chirps in the data provided varied from 18.5 to 44. A prediction at 30 chirps per 15 seconds is inside the domain of our data, so would be interpolation. Using our model: [latex-display]\begin{array}{l}T\left(30\right)=30+1.2\left(30\right)\hfill \\ \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,=66\text{ degrees}\hfill \end{array}[/latex-display] Based on the data we have, this value seems reasonable.
2. The temperature values varied from 52 to 80.5. Predicting the number of chirps at 40 degrees is extrapolation because 40 is outside the range of our data. Using our model: [latex]\begin{array}{l}40=30+1.2c\hfill \\ 10=1.2c\hfill \\ c\approx 8.33\hfill \end{array}[/latex]

We can compare the regions of interpolation and extrapolation using the graph below.

Analysis of the Solution

Our model predicts the crickets would chirp 8.33 times in 15 seconds. While this might be possible, we have no reason to believe our model is valid outside the domain and range. In fact, generally crickets stop chirping altogether below around 50 degrees.

Example: Finding a Least Squares Regression Line

Find the least squares regression line using the cricket-chirp data in the table below.

Chirps	44	35	20.4	33	31	35	18.5	37	26
Temperature	80.5	70.5	57	66	68	72	52	73.5	53

Answer:  

1. Click the plus button (add item) in the upper left corner, and select table.
2. Enter chirps data in the x1 column.
3. Enter temperature data in the y1 column.

   x1	44	35	20.4	33	31	35	18.5	37	26
   y1	80.5	70.5	57	66	68	72	52	73.5	53

4. If you can't see the points on the grid, use the plus and minus buttons in the upper right hand corner to zoom in or out on the grid, or click on the wrench and change the upper bound of x1 to 60, and 100 for the y1.
5. In the empty cell below the table you created, enter the expression y1~mx1+b
6. You can add labels to your graph by clicking on the wrench in the upper right hand corner, and typing them into the cells that say "add a label"

Here is an example of how your graph may look: https://www.desmos.com/calculator/ruvzg6iy3o

Analysis of the Solution

Notice that this line is quite similar to the equation we "eyeballed" but should fit the data better. Notice also that using this equation would change our prediction for the temperature when hearing 30 chirps in 15 seconds from 66 degrees to:

[latex]\begin{array}{l}T\left(30\right)=30.281+1.143\left(30\right)\hfill \\ \text{ }=64.571\hfill \\ \text{ }\approx 64.6\text{ degrees}\hfill \end{array}[/latex]

Example: Finding a Correlation Coefficient

Calculate the correlation coefficient for cricket-chirp data in the table below.

Chirps	44	35	20.4	33	31	35	18.5	37	26
Temperature	80.5	70.5	57	66	68	72	52	73.5	53

Answer: Desmos provides you with the correlation coefficient when you use it to calculate a linear regression. The correlation coefficients is labeled as r = 0.951 for this dataset. This value is very close to 1, which suggests a strong increasing linear relationship. https://www.desmos.com/calculator/ruvzg6iy3o

Example: Using a Regression Line to Make Predictions

Gasoline consumption in the United States has been steadily increasing. Consumption data from 1994 to 2004 is shown in the table below.[footnote]http://www.bts.gov/publications/national_transportation_statistics/2005/html/table_04_10.html[/footnote] Determine whether the trend is linear, and if so, find a model for the data. Use the model to predict the consumption in 2008.Is this an interpolation or an extrapolation?

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

Answer: We can introduce new input variable, t, representing years since 1994, this makes entering the data into Desmos easier. Read the value for b, and the value for the slope, m, from Desmos to create the equation for the regression line:

[latex]C\left(t\right)=113.318+2.209t[/latex]

The correlation coefficient was calculated to be 0.997, suggesting a very strong increasing linear trend. Using this to predict consumption in 2008, which is 14 years after 1994,  [latex]\left(t=14\right)[/latex],

[latex]\begin{array}{l}C\left(14\right)=113.318+2.209\left(14\right)\hfill \\ =144.244\hfill \end{array}[/latex]

The model predicts 144.244 billion gallons of gasoline consumption in 2008. This is an extrapolation because there is not a datapoint whose x1 value is 2008. The scatter plot of the data, including the least squares regression line, is shown below in Desmos. Note how we changed the viewing window for the y-axis to 100 < y < 150. https://www.desmos.com/calculator/lv27pmtdbh