Question

A small town has 2000 inhabitants. At 8 AM, 160 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.)

Answer 1

Answer:

3:36 PM

Step-by-step explanation:

Let fraction (y) of the people that heard the rumor , the differential equation that is satisfied by the by y is,

                            $\frac{dy}{dt} = ky(1-y)$

Solving the differential equation,

                          $y(t) = \frac{y_{o}}{[y_{o} + (1 + y_{o})e^{-kt}]}$

The total number of inhabitants of the town = 2000

The number of people that  heard the rumor = 160

At 8 AM, let t = 0,

               $y(0) = \frac{160}{2000}$

                       = 0.08

By noon, half of the town as heard the rumor.

Then,    

                $y(4) = \frac{1}{2}$

Therefore,

              $\frac{1}{2} = \frac{0.08}{[0.08 + (1 - 0.08)e^{-4k}]}$

              $\frac{0.08}{[0.08 + 0.92e^{-4k}]} = \frac{1}{2}$

              $0.08 + 0.92e^{-4k} = 0.16$  

              $0.92e^{-4k} = 0.16 - 0.08$

              $0.92e^{-4k} = 0.08$

              $e^{-4k} = \frac {0.08}{0.92}$

              $k = - \frac{1}{4} In(\frac {0.08}{0.92})$

              k ≈ 0.06106

Calculating time, t when y(t) = 90%

        ⇒ y(t) = 0.9

              $\frac{0.08}{[0.08 + 0.92e^{-0.06106t}]} =0.9$

              $0.08 + 0.92e^{-0.06106t} = \frac {0.9}{0.08}$

              $0.08 + 0.92e^{-0.06106t} = 0.089$

              $0.92e^{-0.06106t} = 0.089 - 0.08$

              $0.92e^{-0.06106t} = 0.0089$

              $t = - \frac{1}{0.06106}In (\frac{0.0089}{0.92})$

              t = 7.59 hours

              ⇒ 7 hours 36 minutes

From 8 A.M. plus 7 hours 36 minutes = 3:36 PM

At 3:36 PM will 90% of the population have heard the rumor