Starting with an initial value of \( P(0)=25 \), the population of a prairie dog community grows at a rate of \( P^{\prime}(t)=30-\frac{t}{5} \) (in units of prairie dogs/month), for \( 0 \leq t \leq 150 \). a. What is the population 8 months later? b. Find the population \( P(t) \) for \( 0 \leq t \leq 150 \).

To solve for the population of the prairie dog community, we can start by determining the function \( P(t) \). ### Step 1: Find \( P(t) \) We know the growth rate: \[ P'(t) = 30 - \frac{t}{5} \] To get \( P(t) \), we need to integrate \( P'(t) \): \[ P(t) = \int \left( 30 - \frac{t}{5} \right) dt \] This can be split into two terms: \[ P(t) = \int 30 \, dt - \int \frac{t}{5} dt \] Integrating each term gives: \[ P(t) = 30t - \frac{t^2}{10} + C \] Where \( C \) is the constant of integration. Now, we'll use the initial condition \( P(0) = 25 \) to find \( C \): \[ P(0) = 30(0) - \frac{0^2}{10} + C = 25 \] \[ C = 25 \] So, we can write our function for population as: \[ P(t) = 30t - \frac{t^2}{10} + 25 \] ### Step 2: Find \( P(8) \) Now we can find the population 8 months later: \[ P(8) = 30(8) - \frac{8^2}{10} + 25 \] \[ P(8) = 240 - \frac{64}{10} + 25 \] \[ P(8) = 240 - 6.4 + 25 \] \[ P(8) = 258.6 \quad (\text{round to } 259 \text{ prairie dogs}) \] ### Step 3: Population over time for \( 0 \leq t \leq 150 \) Next, we can express \( P(t) \) for \( 0 \leq t \leq 150 \): \[ P(t) = 30t - \frac{t^2}{10} + 25 \] This function gives the population of the prairie dog community over the given time range! **Answer:** a. The population 8 months later is approximately 259 prairie dogs. b. The population \( P(t) \) for \( 0 \leq t \leq 150 \) is given by \( P(t) = 30t - \frac{t^2}{10} + 25 \).