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 module. We can use these values to calculate the slope.

[latex]\begin{array}{l} m=\frac{8100 - 6200}{5 - 0}\hfill \\ \text{}m=\frac{1900}{5}\hfill \\ \text{}m=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]\begin{array}{l}P\left(t\right)=380t+6200 \\ \hfill \end{array}[/latex]

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{}P\left(9\right)=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.

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 triangle.

Using the Pythagorean Theorem, we get:

[latex]\begin{array}{lllllll}D{\left(t\right)}^{2}=A{\left(t\right)}^{2}+E{\left(t\right)}^{2}\hfill & \hfill \\ D{\left(t\right)}^{2}={\left(4t\right)}^{2}+{\left(3t\right)}^{2}\hfill & \hfill \\ D{\left(t\right)}^{2}=16{t}^{2}+9{t}^{2}\hfill & \hfill \\ D{\left(t\right)}^{2}=25{t}^{2}\hfill & \hfill \\ \text{}D\left(t\right)=\pm \sqrt{25{t}^{2}}\hfill & \text{Solve for }D\left(t\right)\text{by taking the square root of each side of the equation}\hfill \\ D{\left(t\right)}=\pm 5t\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.

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}{lllllll}\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{ }y=\frac{1}{3}\left(18\right)\hfill & \hfill \\ \text{ }y=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.

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.

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.

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]\begin{array}{l}T\left(30\right)=30+1.2\left(30\right)\hfill \\ T\left(30\right)=66\text{ degrees}\hfill \end{array}[/latex]
   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 at or below 50 degrees.

1. 1. Enter chirps data in the L1 column.
   2. Enter temperature data in the L2 column.
   3. On a graphing utility, select Linear Regression (LinReg). Using the cricket chirp data from earlier, with technology we obtain the equation [latex]T(c) = 30.281+1.143c[/latex]

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

The graph of the scatter plot with the least squares regression line is shown below:

Scatter plot, showing the line of best fit

Your calculator or software will provide you with the correlation coefficient when you use it to fit 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.

Let t represent years since 1994.

Use your calculator or statistical software 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 \\ C\left(14\right)=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. Note how we changed the viewing window for the y-axis to 100 < y < 150.