The population of a city is growing exponentially. At the start of the year 2000, it had 6 million inhabitants and was growing at a rate of 100 thousand people per year.

During which year will the population size reach 9 million people? (Give a calendar year. For example: 2001)


At what rate will the population be growing at that time? (Give your answer correct to at least three decimal places.)
million people per year

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Answer:

The population will reach 9 million people near the end of 2023.

The population is growing then at the rate of 0.152 million people per year

Step-by-step explanation:

The population of this city can be modeled by the following differential equation.

[tex]P(t) = P_{0}e^{rt}[/tex]

In which [tex]P(t)[/tex] is the population at the time moment t, [tex]P_{0}[/tex] is the initial population and [tex]r[/tex] is the growth rate, decimal.

At the start of the year 2000, it had 6 million inhabitants and was growing at a rate of 100 thousand people per year.

I will say that the year 2000 is [tex]t = 0[/tex].

So [tex]P_{0} = 6,000,000[/tex]

Since it is growing at the rate of 100 thousand people per year, we have that

[tex]P_{1} = 6,000,000 + 100,000 = 6,100,000[/tex].

With this, we can solve the exponential equation and find the value of r.

[tex]P(t) = P_{0}e^{rt}[/tex]

[tex]6,100,000 = 6,000,000e^{r}[/tex]

[tex]e^{r} = 1.017[/tex]

We can apply ln to both sides to find r.

[tex]\ln{e^{r}} = \ln{1.017}[/tex]

[tex]r = 0.017[/tex]

During which year will the population size reach 9 million people?

We find the value of t when [tex]P(t) = 9,000,000[/tex]. Then, t is added to 2000 to find the year. This is because we said that [tex]t = 0[/tex] at the year 2000.

[tex]9,000,000 = 6,000,000e^{0.017t}[/tex]

[tex]e^{0.017t} = 1.5[/tex]

[tex]\ln{e^{0.017t}} = \ln{1.5}[/tex]

[tex]0.017t = 0.405[/tex]

[tex]t = 23.851[/tex]

2000 + 23.851 = 2023.851

The population will reach 9 million people near the end of 2023.

At what rate will the population be growing at that time?

This is the population at the start of 2024 subtracted by the population at the start of 2023.

So:

[tex]P(24) - P(23)[/tex]

[tex]P(24) = 6,000,000e^{0.017*24} = 9,022,843[/tex]

[tex]P(23) = 6,000,000e^{0.017*23} = 8,870,751[/tex]

[tex]P(24) - P(23) = 9,022,843 - 8,870,751 = 152,092[/tex]

The population is growing then at the rate of 152 thousand people a year = 0.152 million people per year

The population size reach 9 million people is 2004.

Exponential function

An exponential function is in the form:

y = abˣ

where y, x are variables, a is the initial value of y and b is the multiplier.

Let y represent the population of the city in millions x years after 2000.

a = 6 million = 6, b = 1 + (100000/1000000) = 0.1 + 1 = 1.1

y = 6(1.1)ˣ

For y = 9:

9 = 6(1.1)ˣ

x = 4.25

The population size reach 9 million people is 2004.

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