Respuesta :
Answer:
a) The differential equation for this model is:
[tex]\dfrac{dy}{dt} =ky(1-y)[/tex]
b) The model y(t) is:
[tex]y(t)=\dfrac{\left(\dfrac{y_0}{1-y_0}\right) e^{kt}}{1+\left(\dfrac{y_0}{1-y_0}\right) e^{kt}}[/tex]
c) t=7.6 hours after 8 AM
Step-by-step explanation:
The rate of spread can be expressed as the first derivative dy/dt, being y the fraction of the population that have heard the rumor.
We know that is proportional to the fraction of the population that have heard the rumor (y) and to the fraction of the population that have not already heard the rumor (y-1). The constant of porportionality is k.
The differential equation becomes then:
[tex]\dfrac{dy}{dt} =ky(1-y)[/tex]
Solving this differential equation:
[tex]\dfrac{dy}{dt} =ky(1-y)\\\\\\ \int \dfrac{dy}{y(1-y)}=k\cdot \int dt \\\\\\ln\left|\dfrac{y}{1-y}\right |=kt+C\\\\\\\dfrac{y}{1-y}=e^{kt+C}=e^Ce^{kt}=Ae^{kt}\\\\\\y=(Ae^{kt})(1-y)\\\\\\y=(Ae^{kt})-(Ae^{kt})y\\\\\\y(1+Ae^{kt})=Ae^{kt}\\\\\\y=\dfrac{Ae^{kt}}{1+Ae^{kt}}[/tex]
The initial condition is y(0)=y0
[tex]y(0)=y_0=\dfrac{Ae^{k*0}}{1+Ae^{k*0}}=\dfrac{A}{1+A}\\\\\\y_0(1+A)=A\\\\\\y_0+y_0A=A\\\\\\A(1-y_0)=y_0\\\\\\A=\dfrac{y_0}{1-y_0}[/tex]
Then the model for y(t) becomes:
[tex]y(t)=\dfrac{\left(\dfrac{y_0}{1-y_0}\right) e^{kt}}{1+\left(\dfrac{y_0}{1-y_0}\right) e^{kt}}[/tex]
c) The initial proportion that have heard the rumor is
[tex]y_0=360/4500=0.08[/tex]
Then we have:
[tex]A=\dfrac{y_0}{1-y_0}=\dfrac{0.08}{1-0.08}=\dfrac{0.08}{0.92}=0.087[/tex]
We also know that at noon (t=4 hours after 8 AM), the fraction y(4) is 0.5.
Then, we can calculate k from there:
[tex]\dfrac{y(4)}{1-y(4)}=Ae^{k(4)}\\\\\\\dfrac{0.5}{0.5}=0.087e^{4k}\\\\\\\dfrac{1}{0.087}=e^{4k}\\\\\\-ln(0.087)=4k\\\\\\k=-ln(0.087)/4=2.44/4=0.61[/tex]
Now we can calculate t, so that y(t)=0.90.
[tex]y=Ae^{kt}(1-y)\\\\\\0.90=0.087e^{0.61t}(1-0.90)=0.10*0.087e^{0.61t}=0.0087e^{0.61t}\\\\\\e^{0.61t}=0.90/0.0087=103.45\\\\\\0.61t=ln(103.45)\\\\\\t=ln(103.45)/0.61=4.64/0.61\\\\\\t=7.60[/tex]
