An urban planner models the population 𝑃(𝑡) (in thousands) of a certain community 𝑡 years from nowby the function 𝑃(𝑡) = 40𝑡𝑡2+10 − 50𝑡+1 + 70a. What is the current population of the community?b. By how much does the population change during the 3rd year? Is the population increasing ordecreasing over this time period?c. What happens to the population in the long run (𝑎𝑠 𝑡 → ∞)
Question
An urban planner models the population 𝑃(𝑡) (in thousands) of a certain community 𝑡 years from nowby the function 𝑃(𝑡) = 40𝑡𝑡2+10 − 50𝑡+1 + 70a. What is the current population of the community?b. By how much does the population change during the 3rd year? Is the population increasing ordecreasing over this time period?c. What happens to the population in the long run (𝑎𝑠 𝑡 → ∞)
Solution
a. The current population of the community can be found by plugging in t=0 into the function P(t). So, P(0) = 40(0)/(0^2+10) - 50(0+1) + 70 = 0 - 50 + 70 = 20 thousand.
b. The change in population during the 3rd year can be found by calculating the difference between the population at the end of the 3rd year and the population at the end of the 2nd year. So, P(3) - P(2) = [40(3)/(3^2+10) - 50(3+1) + 70] - [40(2)/(2^2+10) - 50(2+1) + 70]. Calculate the values inside the brackets first, then subtract the second result from the first to find the change in population. If the result is positive, the population is increasing; if it's negative, the population is decreasing.
c. As t approaches infinity, the term 40t/(t^2+10) will approach 0 because the denominator grows much faster than the numerator. The term -50(t+1) will approach negative infinity, and the term +70 will remain constant. So, as t approaches infinity, the population will approach negative infinity. This means that in the long run, the population will decrease indefinitely.
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