Question

Assume that the carrying capacity for the US population is 800 million. Use it and the fact that the population was 282 million in 2000 to formulate a logistic model for the US population. (Let t = 0 correspond to the year 2000. Use k for your constant.)
(a) P (t) = millions
(b) Determine the value of k in your model by using the fact that the population in 2010 was 309 million. k =
(c) Use your model to predict the US population in the years 2200 and 2400. (Round your answers to the nearest integer.) year 2200 millionyear 2400 million
(d) Use your model to predict the year in which the US population will exceed 500 million. (Round your answer down to the nearest year.)

Answer 1

Answer:

a) $P(t) = \frac{800}{1 + 2.17e^{-kt}}$

b) $k = 0.031$

c) The Population in the year 2200 is 796.76 million.

The population in the year 2400 is 799.99 million.

d) The US population will exceed 500 million in the year of 2041.

Step-by-step explanation:

The logistic equation for population growth is given by:

$P(t) = \frac{C}{1 + \frac{C - P_{0}}{P_{0}}e^{-kt}}$

In which C is the carrying capacity, k is the growth rate, $P_{0}$ is the initial population and $P(t)$ is the population, in millions, after t years.

In this problem, we have that:

Assume that the carrying capacity for the US population is 800 million. This means that [tex}C = 800[/tex].

Use it and the fact that the population was 282 million in 2000 to formulate a logistic model for the US population. In our problem, we consider $t = 0$ being the year 2000. So $P_{0} = 282$. This means that our model is:

a) $P(t) = \frac{C}{1 + \frac{C - P_{0}}{P_{0}}e^{-kt}}$

$P(t) = \frac{800}{1 + \frac{800 - 252}{252}e^{-kt}}$

$P(t) = \frac{800}{1 + 2.17e^{-kt}}$

(b) Determine the value of k in your model by using the fact that the population in 2010 was 309 million.

2010-2000 = 10. This means that $P(10) = 309$.

$P(t) = \frac{800}{1 + 2.17e^{-kt}}$

$309 = \frac{800}{1 + 2.17e^{-10k}}$

$309 + 670.53e^{-10k} = 800$

$670.53e^{-10k} = 491$

$e^{-10k} = 0.73$

Now, we apply ln to both sides to find k

$\ln{e^{-10k}} = \ln{0.73}$

$-10k = -0.31$

$10k = 0.31$

$k = \frac{0.31}}{10}$

$k = 0.031$

(c) Use your model to predict the US population in the years 2200 and 2400.

2200-2000 = 200. So we have to find $P(200)$

$P(t) = \frac{800}{1 + 2.17e^{-0.031t}}$

$P(200) = \frac{800}{1 + 2e^{-0.031(400)}$

$P(200) = 796.76$

The Population in the year 2200 is 796.76 million.

2400 - 2000 = 400. So we have to find $P(400)$.

$P(t) = \frac{800}{1 + 2.17e^{-0.031t}}$

$P(400) = \frac{800}{1 + 2e^{-0.031(400)}$

$P(400) = 799.99$

The population in the year 2400 is 799.99 million.

(d) Use your model to predict the year in which the US population will exceed 500 million.

It is going to be t years after 2000.

This is t when $P(t) = 500$. So:

$P(t) = \frac{800}{1 + 2.17e^{-0.031t}}$

$500 = \frac{800}{1 + 2.17e^{-0.031t}}$

$500 + 1085e^{-0.031t} = 800$

$1085e^{-0.031t} = 300$

$e^{-0.031t} = 0.2765$

Now, we apply ln to both sides.

$\ln{e^{-0.031t}} = \ln{0.2765}$

$-0.031t = -1.2855$

$0.031t = 1.2855$

$t = \frac{1.2855}{0.031}$

$t = 41.46$

2000 + 41.46 = 2041.46. The US population will exceed 500 million in the year of 2041.