Question:

The population of a city is modeled by the function P(t) = 1000 / (1 + 9e^(-0.2t)), where t represents the time in years.

a) Find the equation that represents the rate of change of the population with respect to time.

b) Find the population when t = 5 years.

c) Determine the limiting population (carrying capacity) for this city.

Answer:

a) To find the equation that represents the rate of change of the population with respect to time, we need to differentiate the given function P(t) = 1000 / (1 + 9e^(-0.2t)) with respect to t.

Using the quotient rule, the derivative of P(t) is:

dP/dt = [ (d/dt)(1000) * (1 + 9e^(-0.2t)) - 1000 * (d/dt)(1 + 9e^(-0.2t)) ] / (1 + 9e^(-0.2t))^2

Simplifying the expression, we have:

dP/dt = [ 0 * (1 + 9e^(-0.2t)) - 1000 * (-1) * 9e^(-0.2t) * (-0.2) ] / (1 + 9e^(-0.2t))^2

dP/dt = 1800e^(-0.2t) / (1 + 9e^(-0.2t))^2

Therefore, the equation that represents the rate of change of the population with respect to time is dP/dt = 1800e^(-0.2t) / (1 + 9e^(-0.2t))^2.

b) To find the population when t = 5 years, we substitute t = 5 into the given function P(t):

P(5) = 1000 / (1 + 9e^(-0.2*5)) = 1000 / (1 + 9e^(-1)) ≈ 786.57 (rounded to two decimal places)

Therefore, the population when t = 5 years is approximately 786.57 people.

c) The limiting population (carrying capacity) for this city can be determined by finding the population as t approaches infinity.

As t approaches infinity, the exponential term e^(-0.2t) approaches 0. Therefore, the population approaches:

P(infinity) = 1000 / (1 + 9(0)) = 1000 / 1 = 1000

Therefore, the limiting population (carrying capacity) for this city is 1000 people.

Answer: a) The equation that represents the rate of change of the population with respect to time is dP/dt = 1800e^(-0.2t) / (1 + 9e^(-0.2t))^2.

b) When t = 5 years, the population is approximately 786.57 people.

c) The limiting population (carrying capacity) for this city is 1000 people.