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:

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,
t,
time in hours.
- Output:
A(t),
A(t),
distance in miles, and
E(t),
E(t),
distance in miles

Because it is not obvious how to define our output variable, we’ll start by drawing a picture such as Figure 2.

Initial Value: They both start at the same intersection so when t=0,t=0, 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 AA is 4 and the slope for EE is 3.

Using those values, we can write formulas for the distance each person has walked.

A(t)=4tE(t)=3tA(t)=4tE(t)=3t

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,
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,
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,
D,
to be the measurement of the distance between Anna and Emanuel. Showing the variables on the diagram is often helpful, as we can see from Figure 3.

Recall that we need to know how long it takes for
D,
D,
the distance between them, to equal 2 miles. Notice that for any given input
t,
t,
the outputs
A(
t
),E(
t
),
A(
t
),E(
t
),
and
D(
t
)
D(
t
)
represent distances.

Figure 2 shows us that we can use the Pythagorean Theorem because we have drawn a right angle.

Using the Pythagorean Theorem, we get:

D
(t)
2
=A
(t)
2
+E
(t)
2
=
(4t)
2
+
(3t)
2
=16
t
2
+9
t
2
=25
t
2
D(t)=±
25
t
2
Solve for D(t) using the square root
=±5|t|
D
(t)
2
=A
(t)
2
+E
(t)
2
=
(4t)
2
+
(3t)
2
=16
t
2
+9
t
2
=25
t
2
D(t)=±
25
t
2
Solve for D(t) using the square root
=±5|t|

In this scenario we are considering only positive values of
t,
t,
so our distance
D(
t
)
D(
t
)
will always be positive. We can simplify this answer to
D(t)=5t.
D(t)=5t.
This means that the distance between Anna and Emanuel is also a linear function. Because
D
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
D(t)=2
and solve for
t.
t.

D(t)=2
5t=2
t=
2
5
=0.4
D(t)=2
5t=2
t=
2
5
=0.4

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

AnalysisOne 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.