Question

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

1. Determine the domain: The functions are defined from \( t = 0 \) (the year 2000) to \( t = 20 \) (the year 2020).

2. Evaluate the functions at key points:

   • For \( t = 0 \):
     • \( m(0) = 100(0)\sqrt{3} = 0 \)
     • \( c(0) = 18(0) = 0 \)
   • For \( t = 5 \):
     • \( m(5) = 100(5)\sqrt{3} \approx 100(5)(1.732) \approx 866 \)
     • \( c(5) = 18(5) = 90 \)
   • For \( t = 10 \):
     • \( m(10) = 100(10)\sqrt{3} \approx 1732 \)
     • \( c(10) = 18(10) = 180 \)
   • For \( t = 15 \):
     • \( m(15) = 100(15)\sqrt{3} \approx 2598 \)
     • \( c(15) = 18(15) = 270 \ \)
   • For \( t = 20 \):
     • \( m(20) = 100(20)\sqrt{3} \approx 3464 \)
     • \( c(20) = 18(20) = 360 \)

3. Plot the points on a coordinate grid with \( t \) on the x-axis and the population on the y-axis:

   • \( (0, 0), (5, 866), (10, 1732), (15, 2598), (20, 3464) \) for Mathville.
   • \( (0, 0), (5, 90), (10, 180), (15, 270), (20, 360) \) for Calcfield.

4. Draw the graphs respectively, noting that Mathville's population increases at a much faster rate than Calcfield's.

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B. Point of Intersection:

The two functions intersect at the point where their values are equal, i.e., where \( m(t) = c(t) \). From initial evaluations, \( m(t) \) increases significantly compared to \( c(t) \). Approximation suggests the intersection occurs for small values of \( t \).

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C. Solving Algebraically for Intersection:

To find the intersection, set the two functions equal to each other:

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

Assuming \( t \neq 0 \) (since both functions are zero when \( t = 0 \)), we can divide both sides by \( t \):

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

Now, isolate \( \sqrt{3} \):

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

This equation shows a contradiction, as \( \sqrt{3} \approx 1.732 \). So, it indicates \( t \) cannot yield a positive solution unless both are zero.

The two functions meet only at \( t = 0 \). Thus, the population of both cities starts at 0 in the year 2000.

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D. Comparison of Relative Populations:

Over the time period from 2000 to 2020, Mathville's population grows exponentially compared to Calcfield's linear growth. While both cities start with a population of zero, Mathville's population soars to over 3,000 people by 2020, whereas Calcfield remains under 400 people. This stark contrast illustrates how Mathville's more rapid growth rate positions it well beyond Calcfield's gradual increase.