A Rumor Spreads
Date: 04/01/99 at 22:42:49 From: Rajiv Majithia Subject: Advanced Calculus: Rates 1) A rumor spreads through a community at the rate dy/dt = 2y(1-y), where y is the proportion of the population that has heard the rumor at time t. a) What proportion of the population has heard the rumor when it is spreading the fastest? b) If at time t = 0 ten percent of the people have heard the rumor, find y as a function of t. c) At what time t is the rumor spreading the fastest? Thank you for your help. This has me stumped. I don't know how to even begin to solve this question. Rajiv Majithia
Date: 04/02/99 at 06:43:02 From: Doctor Anthony Subject: Re: Advanced Calculus: Rates >a) What proportion of the population has heard the rumor when it is spreading the fastest? r(y) = dy/dt = 2y(1-y) = 2y - 2y^2 To find y where r(y) is a maximum we differentiate r(y) with respect to y. d(r(y))/dy = 2 - 4y = 0 so y=1/2 So 1/2 the population has heard the rumour when it is spreading fastest. >b) If at time t = 0 ten percent of the people have heard the rumor, find y as a function of t. dy ------- = 2.dt and using partial fractions on the lefthand side y(1-y) [1/y + 1/(1-y)]dy = 2.dt ln(y) - ln(1-y) = 2t + C y ln[-----] = 2t+C 1-y y/(1-y) = e^(2t+C) y/(1-y) = A.e^(2t) at t = 0 y = 0.1 0.1/0.9 = A so A = 1/9 y/(1-y) = (1/9).e^(2t) ...........(1) and this simplifies to: e^(2t) y = ---------- 9+e^(2t) >c) At what time t is the rumor spreading the fastest? We know from part (a) that y = 1/2 when the rumour is spreading fastest, so we put y = 1/2 in above expression and solve for t. In fact it will be best to use equation (1) where the lefthand side will then equal 1. (1/9).e^(2t) = 1 e^(2t) = 9 and taking logs 2t = ln(9) t = ln(9)/2 = 1.0986 units of time (weeks?, months?) - Doctor Anthony, The Math Forum http://mathforum.org/dr.math/
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