Explanation Given: From exercise 31, the solution of the linear differential equation is P ( t ) = M + C e − k t The initial condition P ( 0 ) = 0 If P 1 ( t ) is Jim’s learning curve, then Jim processed 25 units during the first hour, that is P 1 ( 1 ) = 25 Jim processed 45 units during the second hour, that is P 1 ( 2 ) = 45 If P 2 ( t ) is Mark’s learning curve, then Mark processed 35 units during the first hour, that is P 2 ( 1 ) = 35 Mark processed 50 units during the second hour, that is P 2 ( 2 ) = 50 Calculation: The solution of the differential equation is P ( t ) = M + C e − k t (1) Substitute the initial condition P ( 0 ) = 0 in the above equation P ( 0 ) = M + C e − 0 t 0 = M + C e 0 ( ∵ e 0 = 1 ) C + M = 0 C = − M Substitute C = − M in equation (1) P ( t ) = M − M e − k t P ( t ) = M ( 1 − e − k t ) (2) Substitute P 1 ( 1 ) = 25 in equation (2), P 1 ( 1 ) = M 1 ( 1 − e − k ( 1 ) ) 25 = M 1 ( 1 − e − k ) 25 M 1 = 1 − e − k 1 − 25 M 1 = e − k Take natural logarithm on both sides ln ( 1 − 25 M 1 ) = ln ( e − k ) ln ( 1 − 25 M 1 ) = − k − ln ( 1 − 25 M 1 ) = k ln ( M 1 − 25 M 1 ) − 1 = k Simplify further ln ( M 1 M 1 − 25 ) = k Substitute P 1 ( 2 ) = 45 in equation (2) P 1 ( 2 ) = M 1 ( 1 − e − k ( 2 ) ) 45 = M 1 ( 1 − e − 2 k ) Substitute ln ( M 1 M 1 − 25 ) = k in the above equation 45 = M 1 ( 1 − e − 2 ln ( M 1 M 1 − 25 ) ) = M 1 ( 1 − e ln ( M 1 M 1 − 25 ) − 2 ) = M 1 ( 1 − ( M 1 M 1 − 25 ) − 2 ) = M 1 ( 1 − ( M 1 − 25 M 1 ) 2 ) Simplify further, 45 = M 1 ( 1 − ( ( M 1 − 25 ) 2 ( M 1 ) 2 ) ) = M 1 ( 1 − ( ( M 1 2 + 625 − 50 M 1 ) ( M 1 ) 2 ) ) = M 1 ( M 1 2 − M 1 2 − 625 + 50 M 1 ( M 1 ) 2 ) 45 = 50 M 1 − 625 M 1 Simplify further, M 1 45 = 50 M 1 − 625 50 M 1 − M 1 45 = 625 5 M 1 = 625 M 1 = 125 Therefore, the maximum number of units per hour that Jim is capable of processing is M 1 = 125

Student Solutions Manual for Stewart's Single Variable Calculus: Early Transcendentals, 8th (James Stewart Calculus)

8th Edition

ISBN: 9781305272422

Author: James Stewart, Jeffrey A. Cole, Daniel Drucker, Daniel Anderson

Publisher: Cengage Learning

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Chapter 9.5, Problem 32E

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To estimate: The maximum number of units per hour that each worker is capable of processing.

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Chapter 9.5, Problem 31E

Chapter 9.5, Problem 33E

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Student Solutions Manual for Stewart's Single Variable Calculus: Early Transcendentals, 8th (James Stewart Calculus)

Ch. 9.1 - Show that y=23ex+e2x is a solution of the...Ch. 9.1 - Prob. 2E Ch. 9.1 - (a) For what values of r does the function y = erx...Ch. 9.1 - (a) For what values of k does the function y = cos...Ch. 9.1 - Which of the following functions are solutions of...Ch. 9.1 - (a) Show that every member of the family of...Ch. 9.1 - Prob. 7E Ch. 9.1 - Prob. 8E Ch. 9.1 - Prob. 9E Ch. 9.1 - Prob. 10E

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