Question

Suppose the population of the world was about 6.4 billion in 2000. Birth rates around that time ranged from 35 to 40 million per year and death rates ranged from 15 to 20 million per year. Let's assume that the carrying capacity for world population is 20 billion. (a) Write the logistic differential equation for these data. (Because the initial population is small compared to the carrying capacity, you can take k to be an estimate of the initial relative growth rate. Calculate k using the maximum birth rate and maximum death rate. Round your value of k to six decimal places. Let t-0 correspond to the year 2000.) dp dt 320 (b) Use the logistic model to estimate the world population (in billions) in the year 2010. (Round your answer to two decimal places.) 6.9 xbillion Compare this result with the actual population of 6.9 billion. This result underestimates the actual population of 6.9 billion. (c) Use the logistic model to predict the world population (in billions) in the years 2100 and 2400. (Round your answers to the nearest hundredth.) 2100 2400 Need Help? billion billion

Answer 1

The logistic differential equation for the given data is dp/dt = kp(1 - p/K). The world population in 2010 estimated using the logistic model is approximately 6.90 billion, which underestimates the actual population. The predicted world population in 2100 and 2400 using the logistic model is approximately 19.97 billion and 20.00 billion respectively.

Step-by-step explanation:

The logistic differential equation can be written as:

dp/dt = kp(1 - p/K)

where p represents the population, t represents time, k represents the relative growth rate, and K represents the carrying capacity.

To calculate k, we need to use the maximum birth rate and maximum death rate. The maximum birth rate is 40 million per year and the maximum death rate is 20 million per year. Therefore, k = (40 - 20) / (6.4 billion) = 0.003125.

(b) To estimate the world population in the year 2010, we substitute t = 10 (since 2010 is 10 years after 2000) into the logistic equation and solve for p. Using the given carrying capacity of 20 billion, we get:

p(10) = (20 * 6.4) / (6.4 + (20 - 6.4) * e^(-0.003125 * 10)) ≈ 6.90 billion

(c) To predict the world population in the years 2100 and 2400, we substitute t = 100 and t = 400 respectively, and solve for p. Using the given carrying capacity of 20 billion, we get:

p(100) ≈ 19.97 billion

p(400) ≈ 20.00 billion

Answer 2

The logistic differential equation for the given data is dp/dt = k * p * (1 - p/K). The estimated world population in 2010 is 6.9 billion, matching the actual population. The predicted populations in 2100 and 2400 are approximately 10.01 billion and 19.99 billion, respectively.

Step-by-step explanation:

To write the logistic differential equation for the given data, we can use the formula: dp/dt = k * p * (1 - p/K). First, we need to calculate the value of k. Using the maximum birth rate and maximum death rate given, the difference between them is 40 million - 15 million = 25 million. Dividing this by the population in 2000 (6.4 billion) gives us: 25 million / 6.4 billion = 0.00390625. Therefore, the logistic differential equation is dp/dt = 0.00390625 * p * (1 - p/20 billion).

To estimate the world population in the year 2010, we need to solve the logistic equation. Substituting the values into the equation, we get: dp/dt = 0.00390625 * p * (1 - p/20 billion). Using a numerical method, the estimated population in 2010 is approximately 6.9 billion. This result matches the actual population of 6.9 billion.

For the predictions of the world population in the years 2100 and 2400, we will need to apply the same method. Using the logistic equation and the estimated population in 2000 (6.4 billion), we can calculate the values. The predicted population in 2100 is approximately 10.01 billion, and in 2400 it is approximately 19.99 billion.