Respuesta :
Answer:
The differential equation for the model is
[tex]\frac{dP}{dt}=kP(1-P)[/tex]
The model for P is
[tex]P(t)=\frac{1}{1-0.99e^{t/447}}[/tex]
At half day of the 4th day (t=4.488), the population infected reaches 90,000.
Step-by-step explanation:
We can write the rate of spread of the virus as:
[tex]\frac{dP}{dt}=kP(1-P)[/tex]
We know that P(0)=100 and P(3)=100+200=300.
We have to calculate t so that P(t)=0.9*100,000=90,000.
Solving the diferential equation
[tex]\frac{dP}{dt}=kP(1-P)\\\\ \int \frac{dP}{P-P^2} =k\int dt\\\\-ln(1-\frac{1}{P})+C_1=kt\\\\1-\frac{1}{P}=Ce^{-kt}\\\\\frac{1}{P}=1-Ce^{-kt}\\\\P=\frac{1}{1-Ce^{-kt}}[/tex]
[tex]P(0)= \frac{1}{1-Ce^{-kt}}=\frac{1}{1-C}=100\\\\1-C=0.01\\\\C=0.99\\\\\\P(3)= \frac{1}{1-0.99e^{-3k}}=300\\\\1-0.99e^{-3k}=\frac{1}{300}=0.99e^{-3k}=1-1/300=0.997\\\\e^{-3k}=0.997/0.99=1.007\\\\-3k=ln(1.007)=0.007\\\\k=-0.007/3=-0.00224=-1/447[/tex]
Then the model for the population infected at time t is:
[tex]P(t)=\frac{1}{1-0.99e^{t/447}}[/tex]
Now, we can calculate t for P(t)=90,000
[tex]P(t)=\frac{1}{1-0.99e^{t/447}}=90,000\\\\1-0.99e^{t/447}=1/90,000 \\\\0.99e^{t/447}=1-1/90,000=0.999988889\\\\e^{t/447}=1.010089787\\\\ t/447=ln(1.010089787)\\\\t=447ln(1.010089787)=447*0.010039225=4.487533[/tex]
At half day of the 4th day (t=4.488), the population infected reaches 90,000.
