Answer:
The differential equation for the model is
![\frac{dP}{dt}=kP(1-P)](https://tex.z-dn.net/?f=%5Cfrac%7BdP%7D%7Bdt%7D%3DkP%281-P%29)
The model for P is
![P(t)=\frac{1}{1-0.99e^{t/447}}](https://tex.z-dn.net/?f=P%28t%29%3D%5Cfrac%7B1%7D%7B1-0.99e%5E%7Bt%2F447%7D%7D)
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:
![\frac{dP}{dt}=kP(1-P)](https://tex.z-dn.net/?f=%5Cfrac%7BdP%7D%7Bdt%7D%3DkP%281-P%29)
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
![\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}}](https://tex.z-dn.net/?f=%5Cfrac%7BdP%7D%7Bdt%7D%3DkP%281-P%29%5C%5C%5C%5C%20%5Cint%20%5Cfrac%7BdP%7D%7BP-P%5E2%7D%20%3Dk%5Cint%20dt%5C%5C%5C%5C-ln%281-%5Cfrac%7B1%7D%7BP%7D%29%2BC_1%3Dkt%5C%5C%5C%5C1-%5Cfrac%7B1%7D%7BP%7D%3DCe%5E%7B-kt%7D%5C%5C%5C%5C%5Cfrac%7B1%7D%7BP%7D%3D1-Ce%5E%7B-kt%7D%5C%5C%5C%5CP%3D%5Cfrac%7B1%7D%7B1-Ce%5E%7B-kt%7D%7D)
![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](https://tex.z-dn.net/?f=P%280%29%3D%20%20%5Cfrac%7B1%7D%7B1-Ce%5E%7B-kt%7D%7D%3D%5Cfrac%7B1%7D%7B1-C%7D%3D100%5C%5C%5C%5C1-C%3D0.01%5C%5C%5C%5CC%3D0.99%5C%5C%5C%5C%5C%5CP%283%29%3D%20%20%5Cfrac%7B1%7D%7B1-0.99e%5E%7B-3k%7D%7D%3D300%5C%5C%5C%5C1-0.99e%5E%7B-3k%7D%3D%5Cfrac%7B1%7D%7B300%7D%3D0.99e%5E%7B-3k%7D%3D1-1%2F300%3D0.997%5C%5C%5C%5Ce%5E%7B-3k%7D%3D0.997%2F0.99%3D1.007%5C%5C%5C%5C-3k%3Dln%281.007%29%3D0.007%5C%5C%5C%5Ck%3D-0.007%2F3%3D-0.00224%3D-1%2F447)
Then the model for the population infected at time t is:
![P(t)=\frac{1}{1-0.99e^{t/447}}](https://tex.z-dn.net/?f=P%28t%29%3D%5Cfrac%7B1%7D%7B1-0.99e%5E%7Bt%2F447%7D%7D)
Now, we can calculate t for P(t)=90,000
![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](https://tex.z-dn.net/?f=P%28t%29%3D%5Cfrac%7B1%7D%7B1-0.99e%5E%7Bt%2F447%7D%7D%3D90%2C000%5C%5C%5C%5C1-0.99e%5E%7Bt%2F447%7D%3D1%2F90%2C000%20%5C%5C%5C%5C0.99e%5E%7Bt%2F447%7D%3D1-1%2F90%2C000%3D0.999988889%5C%5C%5C%5Ce%5E%7Bt%2F447%7D%3D1.010089787%5C%5C%5C%5C%20t%2F447%3Dln%281.010089787%29%5C%5C%5C%5Ct%3D447ln%281.010089787%29%3D447%2A0.010039225%3D4.487533)
At half day of the 4th day (t=4.488), the population infected reaches 90,000.