Pregunta
upstudy study bank question image url

Assuming \( P \geq 0 \), suppose that a population develops according to the logistic equation \[ \frac{d P}{d t}=0.01 P-5 \times 10^{-5} P^{2} \] where \( t \) is measured in weeks. Answer the following questions. 1. What is the carrying capacity? Carrying Capacity: 2. What is the value of \( k \) ? Answer: \( k= \) Ans 3. For what values of \( P \) is the population increasing? Answer (in interval notation): 4. For what values of \( P \) is the population decreasing? Answer (in interval notation):

Ask by Valdez Potter. in the United States
Mar 14,2025

Solución de inteligencia artificial de Upstudy

Respuesta verificada por el tutor

Responder

**Carrying Capacity:** 200 **Value of \( k \):** 0.01 **Population Increasing:** \(0 < P < 200\) **Population Decreasing:** \(P > 200\)

Solución

¡Inicia sesión para desbloquear respuestas gratis!

Una plataforma de aprendizaje en la que confían millones de estudiantes y profesores reales.

star-icon Descubrir

Respondido por UpStudy AI y revisado por un tutor profesional

Bonus Knowledge

To analyze the logistic growth model given by the equation \[ \frac{d P}{d t} = 0.01 P - 5 \times 10^{-5} P^{2}, \] we can find the answers to the questions as follows: 1. **Carrying Capacity:** The carrying capacity, denoted by \( K \), can be determined from the condition \( \frac{d P}{d t} = 0 \). Setting the equation to zero, we have: \[ 0 = 0.01 P - 5 \times 10^{-5} P^{2}. \] Factoring out \( P \) gives us: \[ P(0.01 - 5 \times 10^{-5} P) = 0. \] This leads to \( P = 0 \) or \( 0.01 - 5 \times 10^{-5} P = 0 \). Solving for \( P \) gives: \[ 0.01 = 5 \times 10^{-5} P \implies P = \frac{0.01}{5 \times 10^{-5}} = 200. \] So, the carrying capacity is \( K = 200 \). 2. **Value of \( k \):** In the logistic equation, \( k \) is defined as the growth rate. Based on the equation \[ \frac{dP}{dt} = rP(1 - \frac{P}{K}), \] we identify \( r = 0.01 \) (the intrinsic growth rate) and find that \( k = 0.01 \). 3. **Values of \( P \) where the population is increasing:** The population is increasing when \( \frac{d P}{d t} > 0 \). From our expression, \[ 0.01 P - 5 \times 10^{-5} P^{2} > 0. \] Factoring yields: \[ P(0.01 - 5 \times 10^{-5} P) > 0. \] This inequality is satisfied when \( P > 0 \) and also when \( 0.01 - 5 \times 10^{-5} P > 0 \), or \( P < 200 \). Thus, the population is increasing in the interval \( (0, 200) \). 4. **Values of \( P \) where the population is decreasing:** The population is decreasing when \( \frac{d P}{d t} < 0 \). Thus, we need to solve \[ 0.01 P - 5 \times 10^{-5} P^{2} < 0. \] This is true when \( P < 0 \) (which we discard since \( P \geq 0 \)) or when \( P > 200 \). Therefore, the population is decreasing for \( (200, \infty) \). In summary: 1. Carrying Capacity: \( 200 \) 2. \( k = 0.01 \) 3. Population increasing: \( (0, 200) \) 4. Population decreasing: \( (200, \infty) \)

Latest Calculus Questions

¡Prueba Premium ahora!
¡Prueba Premium y hazle a Thoth AI preguntas de matemáticas ilimitadas ahora!
Quizas mas tarde Hazte Premium
Estudiar puede ser una verdadera lucha
¿Por qué no estudiarlo en UpStudy?
Seleccione su plan a continuación
Prima

Puedes disfrutar

Empieza ahora
  • Explicaciones paso a paso
  • Tutores expertos en vivo 24/7
  • Número ilimitado de preguntas
  • Sin interrupciones
  • Acceso completo a Respuesta y Solución
  • Acceso completo al chat de PDF, al chat de UpStudy y al chat de navegación
Básico

Totalmente gratis pero limitado

  • Solución limitada
Bienvenido a ¡Estudia ahora!
Inicie sesión para continuar con el recorrido de Thoth AI Chat
Continuar con correo electrónico
O continuar con
Al hacer clic en "Iniciar sesión", acepta nuestros términos y condiciones. Términos de Uso & Política de privacidad