(A) Explanation Given: The epidemic had spread throughout a community of 50,000 people. The rate at which the people get infected is proportional to the product of the number of people who have been infected and the people who have not been infected. The initial number of individuals affected was 100. After 10 days, the number of infected people is 500. Theorem used: Separation of Variables: The solution of the differential equation, f ( y ) y ′ = g ( x ) is given implicitly by the equation, ∫ f ( y ) d y = ∫ g ( x ) d x . Formula used: Logistic growth model: If the rate of change with respect to time t of a quantity y is proportional to both the amount present and the difference between y and a limiting value M , then y satisfies the differential equation, d y d t = k y ( M − y ) , k > 0 , t > 0 Calculation: Let y represents the number of people who have been infected. The total number of people in the community is 50,000. Thus, the number of people who have not been infected will be 50 , 000 − y . The rate is proportional to the product of infected people and uninfected people, which is d y d t = k y ( 50 , 000 − y ) , k > 0 , t > 0 . Thus, the problem follows a logistic growth model. Rewrite the differential equation d y d t = k y ( 50 , 000 − y ) as follows. d y y ( 50 , 000 − y ) = k d t 1 y ( 50 , 000 − y ) d y = k d t The given differential equation d y d t = k y ( 50 , 000 − y ) is in the form of f ( y ) y ′ = g ( x ) where, f ( y ) = 1 y ( 50 , 000 − y ) and g ( t ) = k . By the above mentioned theorem, the solution is given by, ∫ f ( y ) d y = ∫ g ( t ) d t ∫ ( 1 y ( 50 , 000 − y ) ) d y = ∫ k d t (1) Rewrite the integral ∫ ( 1 y ( 50 , 000 − y ) ) d y as, ∫ ( 1 y ( 50 , 000 − y ) ) d y = ∫ ( 50 , 000 + y − y 50 , 000 y ( 50 , 000 − y ) ) d y = 1 50 , 000 ∫ ( 1 y + 1 50 , 000 − y ) d y Obtain the value of each integral as follows. 1 50 , 000 ∫ ( 1 y + 1 50 , 000 − y ) d y = 1 50 , 000 [ ln y − ln ( 50 , 000 − y ) ] Substitute the value of the integral ∫ ( 1 y ( 50 , 000 − y ) ) d y = 1 50 , 000 [ ln y − ln ( 50 , 000 − y ) ] in equation (1) and simplify as follows. ∫ ( 1 y ( 50 , 000 − y ) ) d y = ∫ k d t 1 50 , 000 [ ln y − ln ( 50 , 000 − y ) ] = k t + C 1 ln y − ln ( 50 , 000 − y ) = 50 , 000 ( k t + C 1 ) ln ( y 50 , 000 − y ) = 50 , 000 ( k t + C 1 ) Simplify further as, e ln ( y 50 , 000 − y ) = e 50 , 000 ( k t + C 1 ) y (B) To determine To find: The time taken for the disease to spread around half the community.

Calculus for Business, Economics, Life Sciences, and Social Sciences - Boston U.

15th Edition

ISBN: 9781323047620

Author: Barnett, Ziegler, Byleen

Publisher: Pearson Education

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Chapter 9.2, Problem 77E

(A)

To determine

To find: The number of people who have been infected after 20 days.

(B)

To determine

To find: The time taken for the disease to spread around half the community.

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Chapter 9.2, Problem 76E

Chapter 9.2, Problem 78E

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Calculus for Business, Economics, Life Sciences, and Social Sciences - Boston U.

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The number of people who have been infected after 20 days.

Solution Summary: Theorem used: Logistic growth model: if the rate of change with respect to time t is proportional to both the amount present and the difference between y and a limiting value