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/