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 the rumor and the fraction who have not heard the rumor.

write a differential equation that is satisfied by y

a small town has 3,500 inhabitants. at 8 am, 280 people have heard a rumor. by noon half the town has heard it. at what time will 90% of the population have heard the rumor? (do not round k in your calculation. round your final answer to one decimal place.)

a) dy/dt = ky(1-y)

b) 3:36pm

Step-by-step explanation:

a) Let the number of people who have heard the rumor = p

Let those who have not heard the rumor= q

Total population = p+q

Fraction of those that heard the rumor = p/p+q = y

Fraction of those who did not hear the rumor = q/p+q = 1-y

The rate at which the rumor spreads = dy/dt

dy/dt varies directly to y(1-y)

dy/dt = ky(1-y) where k is a constant

b) Recall that dy/dt = ky(1-y)

y(t) = y/(y+(1-y) e^-kt)

At 8 am , t= 0

y = p/ p+q

y(0) = 280/3500

y(0) = 0.08

By noon(12pm), t = 4

At this time half of the population has heard the rumor

y(4) = 0.5

Recall that y(t) = y/(y+(1-y) e^-kt)

y(t) = y0/(y0+(1-y0) e^-kt)

y(t) = 0.08/(0.08+(1-0.08) e^-kt)

y(t) = 0.08/(0.08+0 92 e^-kt)

To find k, put y(4) = 0.5 into the equation

y(4) = y/(y+(1-y) e^-4k)

0.5 = 0.08/(0.08+0.92e^-4k)

0.08 + 0.92e^-4k = 0.08/0.5

0.92e^-4k = 0.16 - 0.08

0.92e^-4k = 0.08

e^-4k = 0.08/0.92

e^-4k = 0.087

-4k = ln(0.087)

-4k = -2.422

k = -2.422/ -4

k = 0.611

y(t) = 0.08/(0.08+ 0.92 e^-0.611t)

The time by which 90% of the population would have heard the rumor is

0.9 = 0.08/(0.08+0.92e^-0.611t)

0.08 + 0.92e^-0.611t = 0.08/0.9

0.92e^-0.611t= (0.08/0.9) - 0.08

e^-0.611t = [(0.08/0.9)-0.08] / 0.92

e^-0.611t = 0.00966

-0.611t = ln(0.00966)

-0.611t = -4.640

t = -4.640/ -0.611

t = 7.6hrs

t = 7 hrs + (0.6*60)mins

t= 7 hrs + 36mins

t = 7hrs 36 mins

Therefore 8 am + 7 hrs 36 mins = 3:36pm

The time by which 90% of the rumor spreads = 3:36pm