Explanation Given: Number of charter members is 8 , 000 . New members will be attracted at the rate of 200 per month. Calculation: Consider the following function that represents the fraction of people, who will retain their membership in a computer dating service for at least t months, f ( t ) = e − t 10 ...... (1) Number of charter members is equal to 8 , 000 . So out of 8 , 000 members, number of members who will retain their membership after 10 months will be, 8 , 000 f ( 10 ) = 8 , 000 e − 10 10 = 8 , 000 e − 1 New members will be attracted at the rate of 200 per month. Approximate the number of new members who will be active after 10 months, divide the interval 0 ≤ t ≤ 10 into n equal subintervals of length Δ t months. Assume that t j − 1 represents the beginning of j t h sub-interval, so number of new membersaccepted during the j t h sub-interval will be 200 Δ t . 10 months from now, for the j t h sub-interval, remainingmonths will be 10 − t j − 1 . So, out of these 200 Δ t new members, approximately 200 Δ t ⋅ f ( 10 − t j − 1 ) members will retain their membership at that time. So, the total number of new members who will retain their membership 10 months from now can be approximated by the sum ∑ j = 1 n 200 f ( 10 − t j − 1 ) ⋅ Δ t Add the number of current members who will retain their membership after 10 months, total number of members who will retain their membership after 10 months from now will be, P = 8 , 000 e − 1 + ∑ j = 1 n 200 f ( 10 − t j − 1 ) ⋅ Δ t As n increases without bound, the approximation improves and approaches the true value of P

7th Edition

ISBN: 9781524916817

Author: SMITH KARL J, STRAUSS MONTY J, TODA MAGDALENA DANIELE

Publisher: Kendall Hunt Publishing

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Chapter 6.6, Problem 57PS

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Chapter 6 Solutions

Calculus

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The members the service will have ten months from now.

Solution Summary: The author calculates the number of charter members attracted at the rate of 200 per month.

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