AP Calculus AB Exam Question: Logistic Growth

A population follows a logistic growth model given by the equation:

where $P(t)$ represents the population at time $t$, $K$ represents the carrying capacity, $A$ represents the initial condition, and $b$ represents the growth constant.

1. The initial population $P(0)$ is 1000, the carrying capacity $K$ is 5000, and the growth constant $b$ is 0.1. Determine the value of $A$.

Answer:

We are given the values of $P(0)$, $K$, and $b$, and we need to find the value of $A$ in the logistic growth equation.

Given: $P(0) = 1000$, $K = 5000$, and $b = 0.1$.

We know that $P(0) = \frac{K}{1 + A \cdot e^{-b \cdot 0}}$.

Substituting the given values, we have $1000 = \frac{5000}{1 + A \cdot e^{-(0.1) \cdot 0}}$.

Simplifying the equation, we get $1000 = \frac{5000}{1 + A \cdot e^{0}}$.

Since $e^0 = 1$, the equation becomes $1000 = \frac{5000}{1 + A}$.

To find the value of $A$, we can rearrange the equation and solve for $A$.

Multiplying both sides of the equation by $1 + A$, we have $1000(1 + A) = 5000$.

Expanding the equation, we get $1000 + 1000A = 5000$.

Subtracting 1000 from both sides, we have $1000A = 4000$.

Dividing both sides of the equation by 1000, we obtain $A = 4$.

Therefore, the value of $A$ is 4.