Starting with an initial value of \( P(0)=30 \), the population of a prairie dog community grows at a rate of \( P^{\prime}(t)=20-\frac{t}{2} \) (in units of prairie dogs/month), for \( 0 \leq t \leq 40 \). a. What is the population 13 months later? b. Find the population \( P(t) \) for \( 0 \leq t \leq 40 \). a. After 13 months, the population is \( \square \) prairie dogs. (Type a whole number. Round to the nearest prairie dog as needed.)
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To find the population \( P(t) \) for the given situation, we start with the rate of change of the population: \[ P'(t) = 20 - \frac{t}{2} \] We need to integrate this function to find \( P(t) \). 1. **Integrate \( P'(t) \)**: \[ P(t) = \int (20 - \frac{t}{2}) \, dt = 20t - \frac{t^2}{4} + C \] 2. **Determine the constant \( C \)** using the initial condition \( P(0) = 30 \): \[ P(0) = 20(0) - \frac{0^2}{4} + C = 30 \implies C = 30 \] So, we have: \[ P(t) = 20t - \frac{t^2}{4} + 30 \] 3. **Find \( P(t) \) at \( t = 13 \)** months: \[ P(13) = 20(13) - \frac{13^2}{4} + 30 \] Calculating this, \[ P(13) = 260 - \frac{169}{4} + 30 \] \[ = 260 - 42.25 + 30 \] \[ = 260 + 30 - 42.25 \] \[ = 247.75 \] Rounding to the nearest prairie dog gives: \[ P(13) \approx 248 \] Thus, after 13 months, the population is approximately **248** prairie dogs. 4. **Function Summary**: The population function for \( 0 \leq t \leq 40 \) is: \[ P(t) = 20t - \frac{t^2}{4} + 30 \] You can use this function to calculate the prairie dog population at any given time \( t \) within the specified range.