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```
Date: 04/01/99 at 22:42:49
From: Rajiv Majithia

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

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

>a) What proportion of the population has heard the rumor when it is

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/
```
Associated Topics:
High School Calculus

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