The following function gives the population of a town (in thousands of people), t years after it was...

Question

The following function gives the population of a town (in thousands of people), t years after it was founded:
P(t) = 20(1 + e^{10-t})^{-1}
What is the instantaneous rate of change of the population, 6 years after the town was founded?
Choose 1 answer:
A. 0.06 people per year
B. 0.06 thousand people per year
C. 0.35 people per year
D. 0.35 thousand people per year

SnapAndSolve AI · Accepted Answer

To find the instantaneous rate of change of the population at t = 6, we need to find the derivative of P(t) and evaluate it at t = 6.

The function is given by:

P(t) = 20(1 + e^{10-t})^{-1}

Let u = 1 + e^{10-t}, then P(t) = 20u^{-1}.

The derivative of P(t) with respect to t is:

P'(t) = -20u^{-2} * (-e^{10-t})

Substitute u back:

P'(t) = 20e^{10-t} / (1 + e^{10-t})^2

Evaluate P'(t) at t = 6:

P'(6) = 20e^{10-6} / (1 + e^{10-6})^2

Calculate e^{4} and substitute:

P'(6) = 20e^4 / (1 + e^4)^2

Using a calculator, e^4 ≈ 54.598:

P'(6) = 20 * 54.598 / (1 + 54.598)^2

P'(6) = 1091.96 / 3136.36 ≈ 0.35

Therefore, the instantaneous rate of change is approximately 0.35 thousand people per year.

Answer: D. 0.35 thousand people per year

contact us

Contact Info

The following function gives the population of a town (in thousands of people), t years after it was...

Answer

Have a similar question?

More math questions