Spread of a rumor: At what time has 80% heard?

[coke0011]

A rumor spreads in a town according to the following equation:

. . .P = A / (1 + B e^(-Akt))

Suppose 20% of the population hear the rumor at 7am. One hour later, 50% have heard the rumor. At what time has 80% of the population heard the rumor?

The "P" is the number of people, the total population of the town is "A". I think "B" is the number of people that initially hear the rumor. The rate of spread is given by "k", and the time is "t".

I believe the answer is 8:27 AM. Can someone check?

 

[stapel]

Your function is a decreasing one, which doesn't seem to make much sense. Are you sure you have posted it correctly?

When you reply, please confirm or correct your definitions of the variables.

Thank you.

Eliz.

 

[coke0011]

Rumor

Yeah thats the formula our teacher gave us. A is supposed to be the total population of the town. B should be the initial amount of people who hear the rumor, K is the rate of spreading. T is time in hours. I got 0.03 as my K.

 

[tkhunny]

Didn't we do this problem already?

Without additional information, you are only making it up.

A
B
k

Three Variables or Parameters.

7:AM 20%
8:AM 50%

Two hints.

That is no good. You MUST have more information. Were you given the value of k?

If we use 7:AM as t = 0, we have \(\displaystyle \frac{A}{1+B} = \frac{1}{5}\)

This makes 8:AM as t = 1 and we have \(\displaystyle \frac{A}{1+Be^{-Ak}} = \frac{1}{2}\)

I just noticed we are in "Calculus"! Aha! That means we have two more hints. This is a logistic distribution that MUST approach Zero as t decreases without bound and MUST increase toward One as t increases without bound. The negative limit isn't much help, but this mean A = 1. NOW we're in business.

Reviewing:

If we use 7:AM as t = 0, we have \(\displaystyle \frac{1}{1+B} = \frac{1}{5}\)

And B = 4

This makes 8:AM as t = 1 and we have \(\displaystyle \frac{1}{1+4e^{-k}} = \frac{1}{2}\)

And k = ln(4) = 1.38629436

Solving for the inverse function gives: \(\displaystyle t = \frac{1}{ln(4)}*ln(\frac{P*4}{1-P})\)

Without much more difficulty: \(\displaystyle \frac{1}{ln(4)}*ln(\frac{0.8*4}{1-0.8}) = 2\), as in 80% is achieved at exactly 9:AM

Where did you wander off?