# Model of spread of a rumor

From Stewart 7e pg 614 # 9

"One model for the spread of a rumor is that the rate of spread is proportional to the product of the fraction y of the population who have heard th eremor and the fraction who have not hear the rumor. a) Write a differential equation that is satisfied by y. b) Solve the differential equation c) A small town has 1000 inhabitants. At 8 am 80 people have heard a rumor. By noon half the town has heard it. At what will 90 percent of the population have heard the rumor? "

The wording of this is very ambiguous to me and I can't really make sense of it.

They mention a product, so I know that something is being multiplied and that y is a fraction which belongs to the population who have seen it so I think that "have not" heard is a constant, and that y is a fraction taht represents who have. I tried to set this up and it is the wrong answer. I am not sure what they want from that, the English usage is too ambiguous to make sense of it. The complete lack of punctuation is what really does it.

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A start: Let $y=y(t)$ be the fraction who have heard by time $t$. Then the fraction who have not is $1-y$. The rate of change of $y$, we are told, is proportional to the product $y(1-y)$. Our differential equation is therefore $$\frac{dy}{dt}=ky(1-y).$$ This is a special case of the logistic equation, which you know how to solve.

It is convenient to let $t=0$ at $8\colon00$. So $y(0)=\frac{80}{1000}$. We are told that $y(4)=\frac{1}{2}$. These two items are enough to tell us everything about the equation, including the constant $k$. Some algebraic manipulation will be needed.

Now that you have the equation for $y(t)$ in terms of $t$, you can find the $t$ such that $y(t)=0.9$. Note that this $t$ is the time elapsed since $8\colon00$ AM. You will need to give the answer in clock terms.

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I don't understand the use of 1-y to represent of fraction of people who have not heard a rumor. –  user138246 Jun 24 '12 at 21:45
Suppose that the fraction of the town that has heard the rumour is $0.35$, that is, $35\%$. Then the fraction that hasn't heard is $65\%$, that is, $0.65$, which is $1-0.35$. –  André Nicolas Jun 24 '12 at 21:48
Was there any logical way to set this up? I never would have thought of that. –  user138246 Jun 24 '12 at 21:52
Well, you know it now. Thinking about these problems is a way of gathering experience. –  André Nicolas Jun 24 '12 at 21:54

Let $y$ be the fraction of the population that has heard the rumor. The fraction that has not heard the rumor is then $1-y$. Our model would then be $y'=ay(1-y)$ where $a$ is the proportionality constant. Intuitively, the rumor spreads any time somebody who has heard the rumor meets somebody who has not. For a given meeting, the chance that exactly one has heard the rumor is $2y(1-y)$ and the $2$ can be absorbed in the definition ofFor c), your solution from b) has two unknowns: the constant of integration and $a$. You are given two data points, so should be able to determine both constants, then find the time when $y=0.90$ for the final answer.

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