Between 2000 and 2020, the population of Mathville could be modeled by the function m(t)=100t√3

Question

, where m(t)
is the number of people in Mathville, and t
is the number of years since 2000. Between those same years, the population of Calcfield could be modeled by the function c(t)=18t
.

A. Graph each function on graph paper or a neatly made coordinate grid by hand. Be sure to consider an appropriate domain for the functions as you make your graph.

B. Approximately where do the functions intersect? What does this point of intersection represent?

C. Write and solve an equation to algebraically confirm where the two functions intersect. Show your work.

D. Write 2-3 complete sentences comparing the relative populations of the cities over time.

Answer 1

A. Graphing the Functions

To graph the functions \( m(t) = 100t\sqrt{3} \) and \( c(t) = 18t \), we will use the following intervals:

• Domain: Since \( t \) represents the number of years since 2000, we consider \( t \) in the interval [0, 20], corresponding to the years 2000 to 2020.

Here are the values of the functions at key points:

| \( t \) | \( m(t) \) | \( c(t) \) | |-----------|-------------------------------|-----------------| | 0 | \( 100(0)\sqrt{3} = 0 \) | \( 18(0) = 0 \) | | 5 | \( 100(5)\sqrt{3} \approx 865 \) | \( 18(5) = 90 \) | | 10 | \( 100(10)\sqrt{3} \approx 1732 \) | \( 18(10) = 180 \) | | 15 | \( 100(15)\sqrt{3} \approx 2598 \) | \( 18(15) = 270 \) | | 20 | \( 100(20)\sqrt{3} \approx 3464 \) | \( 18(20) = 360 \) |

You can plot these points to create a graph with \( t \) on the x-axis (from 0 to 20) and \( m(t) \) and \( c(t) \) on the y-axis.

B. Point of Intersection

To find the approximate point of intersection, we look for where \( m(t) \) and \( c(t) \) are equal.

By evaluating the functions at different values of \( t \):

• At \( t = 10 \):

  • \( m(10) \approx 1732 \)
  • \( c(10) = 180 \)

• At \( t = 15 \):

  • \( m(15) \approx 2598 \)
  • \( c(15) = 270 \)

It's clear that the two populations do not intersect within 0 to 20 based on these values—\( m(t) \) increases much faster than \( c(t) \).

C. Algebraic Confirmation of Intersection

To find the exact point of intersection, solve \( m(t) = c(t) \):

\[ 100t\sqrt{3} = 18t \]

Assuming \( t \neq 0 \) (as both functions are 0 at \( t=0 \)), we can divide both sides by \( t \):

\[ 100\sqrt{3} = 18 \]

Next, isolate \( t \):

\[ \sqrt{3} = \frac{18}{100} \Rightarrow \sqrt{3} = 0.18 \]

Since \( \sqrt{3} \) is approximately \( 1.732 \), which is not equal to \( 0.18 \), this means that the functions do not intersect for \( t > 0 \).

D. Comparison of Relative Populations

Over the years 2000 to 2020, the population of Mathville, modeled by \( m(t) = 100t\sqrt{3} \), grows significantly faster than the population of Calcfield, modeled by \( c(t) = 18t \). Initially, both cities have the same population of zero in the year 2000. However, as years progress, Mathville's population outstrips Calcfield's dramatically, reaching thousands by the year 2020 while Calcfield remains in the few hundreds. This comparison illustrates a striking disparity in growth rates between the two cities during this period.