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 [latex]y[/latex]-intercept would correspond to the year [latex]0[/latex], more than [latex]2000[/latex] years ago!

To make computation a little nicer, we will define our input as the number of years since 2004:

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

To predict the population in 2013 ([latex]t= 9[/latex]), we would first need an equation for the population. Likewise, to find when the population would reach [latex]15,000[/latex], we would need to solve for the input that would provide an output of [latex]15,000[/latex]. 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 [latex]y[/latex]-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 [latex]y[/latex]-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 [latex]t= 9[/latex].

[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 [latex]9,620[/latex] in 2013.

To find when the population will reach [latex]15,000[/latex], we can set [latex]P\left(t\right)=15000[/latex] and solve for [latex]t[/latex].

[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 [latex]15,000[/latex] in a little more than [latex]23[/latex] 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 [latex]2[/latex] miles apart, which leads us to ask a new question: “How long will it take them to be [latex]2[/latex] miles apart?”
[latex]\\[/latex]
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 [latex]2[/latex] miles apart. We can see that time will be our input variable, so we’ll define our input and output variables.

• Input: [latex]t[/latex], 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 [latex]0[/latex]. Thus the initial value for each is [latex]0[/latex].
[latex]\\[/latex]
Rate of Change: Anna is walking [latex]4[/latex] miles per hour and Emanuel is walking [latex]3[/latex] miles per hour, which are both rates of change. The slope for [latex]A[/latex] is [latex]4[/latex] and the slope for [latex]E[/latex] is [latex]3[/latex].
[latex]\\[/latex]
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, [latex]A[/latex], 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, [latex]E[/latex], to represent Emanuel’s position measured from the starting point in the southward direction.
[latex]\\[/latex]
Note that in defining the coordinate system, we specified both the starting point of the measurement and the direction of measure.
[latex]\\[/latex]
We can then define a third variable, [latex]D[/latex], 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 [latex]D[/latex], the distance between them, to equal [latex]2[/latex] miles. Notice that for any given input [latex]t[/latex], the outputs [latex]A(t)[/latex], [latex]E(t)[/latex], and [latex]D(t)[/latex] 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 [latex]D(t)[/latex] will always be positive. We can simplify this answer to [latex]D(t) = 5t[/latex]. This means that the distance between Anna and Emanuel is also a linear function. Because [latex]D[/latex] is a linear function, we can now answer the question of when the distance between them will reach [latex]2[/latex] miles. We will set the output [latex]D(t) = 2[/latex] and solve for [latex]t[/latex].

[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 [latex]0.4[/latex] hours, or [latex]24[/latex] 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 [latex](30, 10)[/latex], and Eastborough at [latex](20, 0)[/latex].

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 [latex](20, 0)[/latex], 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 [latex](18, 6)[/latex]. 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]

[latex]\\[/latex]
Analysis of the Solution
[latex]\\[/latex]
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.