Explanation Given: The rate at which the paramecia grows is proportional to the product of the number present and the difference between 375 and the number present. The initial number is 15 and reduces to 150 after 1 hour. Formula Used: (1) 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 (2) The integral can be evaluated as follows. ∫ 1 u ( a + b u ) d u = 1 a ln | u ( a + b u ) | 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 . Calculation: From the given condition, it is observed that number of paramecia follows a logistic growth model, which is proportional to the product of the number present and the difference between 375 and the number present. Thus, a logistic growth model will be suitable for the problem. Let P denote the number of paramecia at time t . Then, by the logistic growth model, d P d t = k P ( 375 − P ) . Given P ( 0 ) = 15 and P ( 1 ) = 150 . The aim is to determine the value of P at t = 2 , that is P ( 2 ) . The first step is to use separation of variables to obtain the general solution to the differential equation d P d t = k P ( 375 − P ) . d P P ( 375 − P ) = k d t ∫ 1 P ( 375 − P ) d P = ∫ k d t (1) Compare the integral on the left hand side with the integral in formula ∫ 1 u ( a + b u ) d u = 1 a ln | u ( a + b u ) | . Substitute u = P , a = 375 and b = − 1 in the formula ∫ 1 u ( a + b u ) d u = 1 a ln | u ( a + b u ) | . ∫ 1 P ( 375 − P ) d P = 1 375 ln | P 375 − P | Since P lies in the range 0 < P < 375 , the absolute value signs can be omitted. That is, ∫ 1 P ( 375 − P ) d P = 1 375 ln ( P 375 − P ) . Substitute the value ∫ 1 P ( 375 − P ) d P = 1 375 ln ( P 375 − P ) in equation (1) and simplify as follows. ∫ 1 P ( 375 − P ) d P = ∫ k d t 1 375 ln ( P 375 − P ) = k t + D ln ( P 375 − P ) = 375 k t + 375 D P 375 − P = e 375 k t + 375 D Simplify further as, P 375 − P = e 375 k t e 375 D P 375 − P = e C e B t , where B = 375 k and C = 375 D P = e C e B t ( 375 − P ) P = 375 e C e B t − P e C e B t On further simplification, P + P e C e B t = 375 e C e B t P ( 1 + e C e B t ) = 375 e C e B t P = 375 e C e B t ( 1 + e C e B t ) Multiply numerator and denominator by e − C e − B t

Pearson eText for Calculus for Business, Economics, Life Sciences, and Social Sciences -- Instant Access (Pearson+)

14th Edition

ISBN: 9780137554805

Author: Raymond Barnett, Michael Ziegler

Publisher: PEARSON+

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The number of paramecia present after 2 hours.

Solution Summary: The author explains that number of paramecia follows a logistic growth model, which is proportional to the amount present and the difference between y and limiting value M.

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