In 2012, the population of a city was 6.57 million. The exponential growth rate was 2.87% per year. a) Find the exponential growth function. b) Estimate the population of the city in 2018. c) When will the population of the city be 10 million? d) Find the doubling time. a) The exponential growth function is P(t)equals nothing, where t is in terms of the number of years since 2012 and P(t) is the population in millions
Accepted Solution
A:
Answer:a) [tex]P(t) = 6.57e ^{0.0287t}[/tex]b) [tex]P(6) = 7,805\ million[/tex]c) [tex]t = 14.64\ years[/tex]d) [tex]t = 24.15\ years[/tex]Step-by-step explanation:
a) The function of exponential growth for a population has the following formula:
[tex]P(t) = p_0e ^{rt}[/tex]In this equation:
[tex]p_0[/tex] is the initial population
r is the growth rate
t is the time in years
In this problem we know that
[tex]r = 2.87\% = 0.0287[/tex][tex]p_0= 6.57[/tex] million in the year 2012.
So the equation is:
[tex]P(t) = 6.57e ^{0.0287t}[/tex]Where [tex]t = 0[/tex] represents the year 2012
b) If [tex]t = 0[/tex] in 2012, then in 2018 [tex]t = 6[/tex]The population in 2018 is:
[tex]P(t = 6) = 6.57e ^{0.0287(6)}\\\\P(6) = 7,805\ million[/tex]
c) To know when the population is equal to 10 million we must equal P(t) to 10 and solve for t.
[tex]P(t) = 10 = 6.57e ^{0.0287t}\\\\\frac{10}{6.57} = e ^{0.0287t}\\\\ln(\frac{10}{6.57}) = 0.0287t\\\\t = \frac{ln(\frac{10}{6.57})}{0.0287}[/tex][tex]t = 14.64\ years[/tex]
d) The function is doubled when [tex]P(t) = 2p_0[/tex][tex]P(t) = 2(6.57) = 13.14 = 6.57e ^{0.0287t}[/tex]We solve for t.
[tex]\frac{13.14}{6.57} = e ^{0.0287t}\\\\ln(\frac{13.14}{6.57}) = 0.0287t\\\\t = \frac{ln(\frac{13.14}{6.57})}{0.0287})[/tex][tex]t = 24.15\ years[/tex]