Pregunta
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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.)

Ask by Ray Mckenzie. in the United States
Mar 20,2025

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a. After 13 months, the population is 248 prairie dogs.

Solución

We begin with the differential equation for the growth rate of the population: \[ P'(t)=20-\frac{t}{2}, \] and the initial condition: \[ P(0)=30. \] ### Part (a) To find the population after 13 months, we integrate the growth rate from \( t=0 \) to \( t=13 \) and add the initial population: \[ P(13)=P(0)+\int_{0}^{13}\left(20-\frac{t}{2}\right)dt. \] First, we compute the integral: \[ \int_{0}^{13}\left(20-\frac{t}{2}\right)dt = \int_{0}^{13}20\,dt-\int_{0}^{13}\frac{t}{2}\,dt. \] Evaluate the first integral: \[ \int_{0}^{13}20\,dt=20t\Big|_{0}^{13}=20(13)=260. \] Next, evaluate the second integral: \[ \int_{0}^{13}\frac{t}{2}\,dt=\frac{1}{2}\int_{0}^{13}t\,dt=\frac{1}{2}\left(\frac{t^2}{2}\Big|_{0}^{13}\right)=\frac{1}{2}\left(\frac{169}{2}\right)=\frac{169}{4}=42.25. \] Subtracting, we have: \[ \int_{0}^{13}\left(20-\frac{t}{2}\right)dt=260-42.25=217.75. \] Thus, the population at \( t=13 \) is: \[ P(13)=30+217.75=247.75. \] Rounding to the nearest whole number, we have: \[ P(13)=248 \quad \text{prairie dogs}. \] ### Part (b) To find the general population function \( P(t) \), we integrate \( P'(t) \): \[ P(t)=\int\left(20-\frac{t}{2}\right)dt. \] Integrate term-by-term: \[ \int 20\,dt=20t, \] \[ \int \frac{t}{2}dt=\frac{1}{2}\cdot\frac{t^2}{2}=\frac{t^2}{4}. \] So, the antiderivative is: \[ P(t)=20t-\frac{t^2}{4}+C, \] where \( C \) is the constant of integration. Using the initial condition \( P(0)=30 \): \[ 30=20(0)-\frac{0^2}{4}+C \quad \Longrightarrow \quad C=30. \] Thus, the population function is: \[ P(t)=20t-\frac{t^2}{4}+30, \quad 0\leq t\leq40. \] --- a. After 13 months, the population is \(248\) prairie dogs.

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The Deep Dive

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.

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