Starting with an initial value of P(0) = 50, the population of a prairie dog community grows at a rate of P'()= 10- (in units of prairie dogs/month), for Osts50. a. What is the population 12 months later? b. Find the population P(t) for 0sts50.
Starting with an initial value of P(0) = 50, the population of a prairie dog community grows at a rate of P'()= 10- (in units of prairie dogs/month), for Osts50. a. What is the population 12 months later? b. Find the population P(t) for 0sts50.
Calculus: Early Transcendentals
8th Edition
ISBN:9781285741550
Author:James Stewart
Publisher:James Stewart
Chapter1: Functions And Models
Section: Chapter Questions
Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...
Related questions
Question
![### Prairie Dog Population Growth Analysis
#### Introduction
We are given a model to understand the growth dynamics of a prairie dog community starting with an initial population of 50 prairie dogs. The growth rate of the population is described by the following differential equation:
\[ P'(t) = 10 - \frac{t}{5} \]
This growth rate is provided in units of prairie dogs per month, applicable over the time interval \(0 \leq t \leq 50\) months.
#### Problem Analysis
**a. What is the population 12 months later?**
To determine the population 12 months from the initial time (t=0), we integrate the given differential equation and apply the initial condition \(P(0) = 50\).
**b. Find the population \(P(t)\) for \(0 \leq t \leq 50\).**
We need to integrate the population growth rate over the given interval to find an explicit function for \(P(t)\).
#### Solution Steps
1. **Integration of the Differential Equation:**
\[ P'(t) = 10 - \frac{t}{5} \]
Integrating both sides with respect to \(t\):
\[ \int P'(t) \, dt = \int \left( 10 - \frac{t}{5} \right) dt \]
\[ P(t) = 10t - \frac{t^2}{10} + C \]
2. **Apply the Initial Condition:**
Given \(P(0) = 50\),
\[ 50 = 10(0) - \frac{0^2}{10} + C \]
\[ C = 50 \]
Thus, the population function is:
\[ P(t) = 10t - \frac{t^2}{10} + 50 \]
3. **Population at t = 12 (12 months later):**
\[ P(12) = 10(12) - \frac{12^2}{10} + 50 \]
\[ P(12) = 120 - \frac{144}{10} + 50 \]
\[ P(12) = 120 - 14.4 + 50 \]
\[ P(12) = 155.6 \]
Therefore, the population 12 months later is approximately 155.6 prairie dogs.
4. **Population over \(0 \leq](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F777e0514-8160-4940-b57c-96884b4e2915%2F64181f59-6925-471e-aee4-2e9681dcd04b%2Ff4lj2c_processed.png&w=3840&q=75)
Transcribed Image Text:### Prairie Dog Population Growth Analysis
#### Introduction
We are given a model to understand the growth dynamics of a prairie dog community starting with an initial population of 50 prairie dogs. The growth rate of the population is described by the following differential equation:
\[ P'(t) = 10 - \frac{t}{5} \]
This growth rate is provided in units of prairie dogs per month, applicable over the time interval \(0 \leq t \leq 50\) months.
#### Problem Analysis
**a. What is the population 12 months later?**
To determine the population 12 months from the initial time (t=0), we integrate the given differential equation and apply the initial condition \(P(0) = 50\).
**b. Find the population \(P(t)\) for \(0 \leq t \leq 50\).**
We need to integrate the population growth rate over the given interval to find an explicit function for \(P(t)\).
#### Solution Steps
1. **Integration of the Differential Equation:**
\[ P'(t) = 10 - \frac{t}{5} \]
Integrating both sides with respect to \(t\):
\[ \int P'(t) \, dt = \int \left( 10 - \frac{t}{5} \right) dt \]
\[ P(t) = 10t - \frac{t^2}{10} + C \]
2. **Apply the Initial Condition:**
Given \(P(0) = 50\),
\[ 50 = 10(0) - \frac{0^2}{10} + C \]
\[ C = 50 \]
Thus, the population function is:
\[ P(t) = 10t - \frac{t^2}{10} + 50 \]
3. **Population at t = 12 (12 months later):**
\[ P(12) = 10(12) - \frac{12^2}{10} + 50 \]
\[ P(12) = 120 - \frac{144}{10} + 50 \]
\[ P(12) = 120 - 14.4 + 50 \]
\[ P(12) = 155.6 \]
Therefore, the population 12 months later is approximately 155.6 prairie dogs.
4. **Population over \(0 \leq
Expert Solution
This question has been solved!
Explore an expertly crafted, step-by-step solution for a thorough understanding of key concepts.
Step by step
Solved in 2 steps with 1 images
Knowledge Booster
Learn more about
Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, calculus and related others by exploring similar questions and additional content below.Recommended textbooks for you
Calculus: Early Transcendentals
Calculus
ISBN:
9781285741550
Author:
James Stewart
Publisher:
Cengage Learning
Thomas' Calculus (14th Edition)
Calculus
ISBN:
9780134438986
Author:
Joel R. Hass, Christopher E. Heil, Maurice D. Weir
Publisher:
PEARSON
Calculus: Early Transcendentals (3rd Edition)
Calculus
ISBN:
9780134763644
Author:
William L. Briggs, Lyle Cochran, Bernard Gillett, Eric Schulz
Publisher:
PEARSON
Calculus: Early Transcendentals
Calculus
ISBN:
9781285741550
Author:
James Stewart
Publisher:
Cengage Learning
Thomas' Calculus (14th Edition)
Calculus
ISBN:
9780134438986
Author:
Joel R. Hass, Christopher E. Heil, Maurice D. Weir
Publisher:
PEARSON
Calculus: Early Transcendentals (3rd Edition)
Calculus
ISBN:
9780134763644
Author:
William L. Briggs, Lyle Cochran, Bernard Gillett, Eric Schulz
Publisher:
PEARSON
Calculus: Early Transcendentals
Calculus
ISBN:
9781319050740
Author:
Jon Rogawski, Colin Adams, Robert Franzosa
Publisher:
W. H. Freeman
Calculus: Early Transcendental Functions
Calculus
ISBN:
9781337552516
Author:
Ron Larson, Bruce H. Edwards
Publisher:
Cengage Learning