(a) Explanation Given information: A pond on a fish farm carrying capacity 2000 fish and the pond was originally stocked with 100 fish. The function N ( t ) is the number of fish in the pond after t months. Explanation: The logistic equation is of the form d N d t = r K N ( K − N ) . Here, the carrying capacity is K = 2000 . Therefore, the logistic differential equation is d N d t = r 2000 N ( 2000 − N ) . The pond was originally stocked with 100 fish. Hence, the initial value of N is N ( 0 ) = 100 . Now, the constant solutions are the solution of the equation d N d t = r 2000 N ( 2000 − N ) = 0 . ⇒ r 2000 N ( 2000 − N ) = 0 ⇒ N ( 2000 − N ) = 0 ⇒ N = 0 or N = 2000 . Now, the inflection points are the points where d 2 N d t 2 = 0 Therefore, differentiate the given differential equation with respect to t , ⇒ d d t ( d N d t ) = d d t ( r N − r 2000 N 2 ) ⇒ d 2 N d t 2 = r d N d t − r 2000 &# (b) To determine To calculate: The size of the population of fish with the highest rate of growth and find the rate when the intrinsic rate is 0.3 .

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ISBN: 9780137590728

Author: Larry Goldstein, David Lay

Publisher: PEARSON+

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Chapter 10.6, Problem 6E

(a)

To determine

The logistic differential equation satisfied by N(t), where N(t) is the number of fish in the pond after t months and plot the approximate graph of the fish population.

(b)

To determine

To calculate: The size of the population of fish with the highest rate of growth and find the rate when the intrinsic rate is 0.3.

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Chapter 10.6, Problem 5E

Chapter 10.6, Problem 7E

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The logistic differential equation satisfied by N ( t ) , where N ( t ) is the number of fish in the pond after t months and plot the approximate graph of the fish population.

Solution Summary: The author explains the logistic differential equation satisfied by N(t) and plots the approximate graph of the fish population.