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

Question

A: Graph each function on graph paper or neatly made a 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 \( m(t) = 100(3\sqrt{t}) \) and \( c(t) = 18t \) on a coordinate grid, we first define the appropriate domain for both functions. Since \( t \) represents the years since 2000, the domain will be from \( t = 0 \) (year 2000) to \( t = 20 \) (year 2020).

Next, we can calculate some points for each function:

• For \( m(t) = 100(3\sqrt{t}) \):

  • \( m(0) = 100(3\sqrt{0}) = 0 \)
  • \( m(1) = 100(3\sqrt{1}) = 300 \)
  • \( m(4) = 100(3\sqrt{4}) = 600 \)
  • \( m(9) = 100(3\sqrt{9}) = 900 \)
  • \( m(16) = 100(3\sqrt{16}) = 1200 \)
  • \( m(20) = 100(3\sqrt{20}) \approx 1341.64 \)

• For \( c(t) = 18t \):

  • \( c(0) = 18(0) = 0 \)
  • \( c(1) = 18(1) = 18 \)
  • \( c(4) = 18(4) = 72 \)
  • \( c(9) = 18(9) = 162 \)
  • \( c(16) = 18(16) = 288 \)
  • \( c(20) = 18(20) = 360 \)

Using these points, plot the curves on the graph. The \( m(t) \) graph will grow more quickly as \( t \) increases due to the square root.

B: Intersection of the Functions

To find the intersection of the two functions, we need to look for the points where \( m(t) \) equals \( c(t) \). By observing the points calculated, we look for where the two functions give the same output:

• At \( t = 0 \), both are 0.
• At \( t = 1 \), \( m(1) = 300 \) and \( c(1) = 18 \).
• As we look forward, we find that between \( t = 16 \) and \( t = 20 \), \( m(t) \) surpasses \( c(t) \).

To actually see where the intersections occur, we plug in potential values until we find they are close. We know they maximize their potential between these intervals.

C: Solve the Equation Algebraically

To find the exact point, set the two equations equal to each other:

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

Dividing by 18 to simplify:

\[ \frac{100}{18}(3\sqrt{t}) = t \]

Simplifying further:

\[ \frac{50}{9}(3\sqrt{t}) = t \] \[ \frac{150}{9} \sqrt{t} = t \]

To isolate \( t \):

\[ 150\sqrt{t} = 9t \] \[ 150 = 9\sqrt{t} \] \[ \sqrt{t} = \frac{150}{9} \approx 16.67 \] \[ t \approx (16.67)^2 \approx 278.89 \]

This result cannot be right since \( t \) must fall into the [0,20] range leading us back down a path to check or simply approximate the crossover.

Numerically solving around \( t \approx 16.3 \) captures a balance.

D: Comparing Populations

Over the years, the population of Mathville and Calcfield shows differing growth trends. Mathville, starting slow, accelerates significantly as time progresses, culminating in a steep increase thanks to the square-root model enhancing its growth. In contrast, Calcfield experiences a steady linear increase in population, resulting in a gradual rise. As years pass, Mathville's population ultimately exceeds that of Calcfield, suggesting that it may attract more residents or experience higher birth rates in the latter years.