A rumor spreads through a school. Let y (t) be the fraction of the population th

Question

A rumor spreads through a school. Let y (t) be the fraction of the population that has heard the rumor at time t and assume that the rate at which the rumor spreads is proportional to the product of the fraction y of the population that has heard the rumor and the fraction 1 - y that has not yet heard the rumor. The school has 1000 students in total. At 8 a.m., 104 students have heard the rumor, and by noon, half the school has heard it. Using the logistic model explained above, determine how much time passes before 90% of the students will have heard the rumor. 90% of the students have heard the rumor after about hours.

Explanation / Answer

Solution:

The rate at which the spreads is proportional to the product of people heard the rumor and people not heard the rumor.
That is,
dy/dt y(1 - y)
Let k be proportionality constant
then dy/dt = ky(1-y)
On solving
y(t) = 1/1+Ce^-kt
To evalaute C, plug in 104 for y and 0 for t in the above equation

104 = 1/1+ Ce^-k(0)
104 = 1/1+C
C = -104/105
Therefore
y(t) = 1/ 1- 104/105 e^-kt
At noon, t = 4 since, t = 0 represents 8 a.m
plug in 500 for y and 4 for t:
500 = 1/ 1-104/105 e^-4k
k = -0.0018
then y(t) = 1/ 1-104/105 e^-0.0018t

The time taken so that the rumor spreads to 90% of people is
900 = 1/(1-(104/105)) e^-0.0018t
e^-0.0018t = 7/60
t = 0.11688
So, it takes approximately 1.16 hrs to spread the rumor

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.