(a) To determine The maximum population of the predator. Answer The maximum population of the predators is 8900 . Explanation Given: The function for the predator’s population is, y = 900 cos 2 t + 8000 (1) Calculation: The equation for the simple harmonic motion which describes the displacement y of an object at time t is, y = a cos ( ω ( t − c ) ) + b (2) For the maximum displacement the cosine function should be maximum. So, the cosine function is maximum at t = 0 . Substitute 0 for t in equation (1), y = 900 cos 2 ⋅ 0 + 8000 = 900 cos 0 + 8000 [ ∴ cos 0 = 1 ] = 900 + 8000 = 8900 Thus, the maximum population of the predators is 8900. (b) To determine The time between the successive periods of maximum population. Answer The time between the successive intervals of maximum displacement is 3.14 years. Explanation Given: The function for the predator’s population is, y = 900 cos 2 t + 8000 (1) Calculation: The equation for the simple harmonic motion which describes the displacement y of an object at time t is, y = a cos ( ω ( t − c ) ) + b (2) Calculate time period for the maximum displacement, For the maximum displacement the cosine value should be 0. So, the value is, 2 t = 0 (3) For the cosine curve the time period is, p = 2 π (4) Calculate time period for the maximum displacement, Equate the equation (3) and (4), 2 t = 2 π t = π t ≈ 3.14 years Thus, the length of time for the maximum displacement between successive period is 3.14 years .

Precalculus: Mathematics for Calculus - 6th Edition

6th Edition

ISBN: 9780840068071

Author: Stewart, James, Redlin, Lothar, Watson, Saleem

Publisher: Cengage Learning

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Textbook Question

Chapter 5.6, Problem 30E

Predator Population Model In a predator/prey model, the predator population is modeled by the function

y = 900 cos 2 t + 8000

where t is measured in years.

(a) What is the maximum population?
(b) Find the length of time between successive periods of maximum population.

(a)

Expert Solution

To determine

The maximum population of the predator.

Answer to Problem 30E

The maximum population of the predators is 8900 .

Explanation of Solution

Given:

The function for the predator’s population is,

y=900cos2t+8000 (1)

Calculation:

The equation for the simple harmonic motion which describes the displacement y of an object at time t is,

y=acos(ω(t−c))+b (2)

For the maximum displacement the cosine function should be maximum. So, the cosine function is maximum at t=0 .

Substitute 0 for t in equation (1),

y=900cos2⋅0+8000=900cos0+8000 [∴cos0=1]=900+8000=8900

Thus, the maximum population of the predators is 8900.

(b)

Expert Solution

To determine

The time between the successive periods of maximum population.

Answer to Problem 30E

The time between the successive intervals of maximum displacement is 3.14 years.

Explanation of Solution

Given:

The function for the predator’s population is,

y=900cos2t+8000 (1)

Calculation:

The equation for the simple harmonic motion which describes the displacement y of an object at time t is,

y=acos(ω(t−c))+b (2)

Calculate time period for the maximum displacement,

For the maximum displacement the cosine value should be 0. So, the value is,

2t=0 (3)

For the cosine curve the time period is,

p=2π (4)

Calculate time period for the maximum displacement,

Equate the equation (3) and (4),

2t=2πt=πt≈3.14 years

Thus, the length of time for the maximum displacement between successive period is 3.14 years .

Chapter 5.6, Problem 29E

Chapter 5.6, Problem 31E

Chapter 5 Solutions

Precalculus: Mathematics for Calculus - 6th Edition

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Predator Population Model In a predator/prey model, the predator population is modeled by the function y = 900 cos 2 t + 8000 where t is measured in years. (a) What is the maximum population? (b) Find the length of time between successive periods of maximum population.

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Answer to Problem 30E

Explanation of Solution

Answer to Problem 30E

Explanation of Solution

Chapter 5 Solutions