MTH133 - JustAnswer
2) Decide on an initial population, [pic], of the town in the year 2010. Choose an initial population between 1000–5000. Use this value of [pic] for each of the scenarios. P0 = __1000_________
3) You will investigate four different scenarios of population growth or decline in this town.
• Linear growth
• Growth modeled by a quadratic equation
• Growth modeled by a radical equation
• Population decline modeled by a rational equation
I. Linear Growth:
Suppose that the amount that your town’s population grows each year is fixed (or constant).
Choose the amount of population growth each year = ___200____
(Hint: Choose a whole number for your growth rate, rather than a percent.)
a) Fill in the following chart:
|Year (t) |Population (P) |
|t = 0 |[pic]1000 |
|(2010) | |
|t = 1 |1000 + 200 = 1200 |
|(2011) | |
|t = 2 |1000 + 400 = 1400 |
|(2012) | |
|t = 3 |1000 + 600 = 1600 |
|(2013) | |
|t = 6 |1000 + 1200 = 2200 |
|(2016) | |
b) Find a linear equation in the form P = mt + b (y = mx + b), which gives the population, P, t years from 2010.
Answer: P = 200t + 1000
Show your work here:
Using slope intercept form, with slope = growth rate = 200, intercept = initial population = 1000
c) Use your equation in part b to approximate the population in the year 2030.
Answer: 5000
Show your work here:
Set t = 20:
200*20 + 1000
= 4000 + 1000
= 5000
d) Use your equation in part b to approximate how many years it will take the population to reach 7000.
Answer: 30 years
Show your work here:
Set P = 7000:
7000 = 200t + 1000
Subtract 1000:
6000 = 200t
Divide by 200:
30 = t
e) Graph this function in MS Excel by plotting the points found in your chart in part a. Label your axes with time on the x-axis and population on the y-axis. Copy and paste your graph here:
Answer:
[pic]
II. Quadratic Growth:
Suppose instead that the town experiences quadratic growth of the form
[pic] where t is the time in years from 2010.
a) Insert the value of [pic]that your group has decided upon into the equation. Use t^2 to type t-squared.
Answer: 8t^2 + 1000
b) Fill in the following chart.
|Year (t) |Population (P) |
|t = 0 |[pic]1000 |
|(2010) | |
|t = 1 |8+1000 = 1008 |
|(2011) | |
|t = 2 |8*4+1000 = 1032 |
|(2012) | |
|t = 3 |8*9+1000 = 1072 |
|(2013) | |
|t = 6 |8*36+1000 = 1288 |
|(2016) | |
c) Use your equation from part a to approximate how many years it will take for the population to reach 7000. Round to the nearest whole year when necessary.
Answer: 27.4 years
Show your work here:
Set P = 7000:
7000 = 8t^2 + 1000
Subtract 1000:
6000 = 8t^2
Divide by 8:
750 = t^2
Square root:
27.4 = t
d) Graph this function in MS Excel by plotting the points found in your chart in part b. Label your axes with time on the x-axis and population on the y-axis. Copy and paste your graph here:
Answer:
[pic]
III. Growth Modeled by a Radical Equation:
Suppose instead that the town experiences growth that can be modeled by the following: [pic] where t is the number of years from 2010.
a) Insert the value of [pic]that your group has decided upon into the equation above. Use the Equation Editor or type square root of t as sqrt(t).
Answer: 1000sqrt(t) + 1000
b) Fill in the following chart. Round to the nearest whole person when necessary.
|Year (t) |Population (P) |
|t = 0 |[pic]1000 |
|(2010) | |
|t = 1 |1000 + 1000 = 2000 |
|(2011) | |
|t = 2 |1414+1000 = 2414 |
|(2012) | |
|t = 3 |1732+1000 = 2732 |
|(2013) | |
|t = 6 |2449+1000 = 3449 |
|(2016) | |
c) Use your equation from part a to approximate how many years it would take for the population to reach 7000. Round to the nearest whole year when necessary.
Answer: 36 years
Show your work here:
Set P = 7000:
7000 = 1000sqrt(t) + 1000
Subtract 1000:
6000 = 1000sqrt(t)
Divide by 1000:
6 = sqrt(t)
Square it:
36 = t
d) Graph this function in MS Excel by plotting the points found in your chart in part b. Label your axes with time on the x-axis and population on the y-axis. Copy and paste your graph here:
Answer:
[pic]
IV. Population Decline Modeled by a Rational Equation:
Suppose instead that the town experiences population decline that can be modeled by the following:[pic] where t is the number of years from 2010.
a) Insert the value of [pic]that your group has agreed to use.
Type as ([pic]) / (t + 1) or use the Equation Editor.
Answer: 1000/(t+1)
b) Fill in the following chart. Round to the nearest whole person when necessary.
|Year (t) |Population (P) |
|t = 0 |[pic]1000 |
|(2010) | |
|t = 1 |1000/2 = 500 |
|(2011) | |
|t = 2 |1000/3 = 333 |
|(2012) | |
|t = 3 |1000/4 = 250 |
|(2013) | |
|t = 6 |1000/7 = 143 |
|(2016) | |
c) Use your equation from part a to approximate how many years it would take for the population to reach 600. Round to two decimal places if necessary.
Answer: 0.67 years
Show your work here:
Set P = 600:
600 = 1000/(t+1)
Multiply by t+1:
600(t+1) = 1000
Divide by 600:
(t+1) = 5/3
Subtract 1:
t = 2/3
t = 0.67
d) Graph this function in MS Excel by plotting the points in the chart in part b. Label your axes with time on the x-axis and population on the y-axis. Copy and paste graph here:
Answer:
[pic]
V.
Suppose that the mayor of the town you have chosen has built a new factory in hopes of drawing as many new people to the town as possible. Which of the four models would the mayor hope that the population would follow? Explain.
If I was the mayor of the town, I would want the population to follow the quadratic model. This model starts out by growing slowly, giving the town a chance to expand and build new roads and schools. Then the growth rate picks up, and the population reaches 7000 before any of the other models. I certainly would not want the rational model, since the population goes down in that model. The radical equation model grows too quickly at the start, and then stops growing so fast, while the linear model grows at the same rate over time. Therefore, the quadratic model is the best one.
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