Question

mondays lecture went over a logistic equation: $f(t)=\frac{1000}{1 + 999e^{-0.603t}}$ which models how many people $f$ have had the flu $t$ days after an initial infection in a community of 1000 people. according to the model, how many people will have had the flu after 10 days? according to the model, how many people will have had the flu after 20 days? according to the model, how many days until 500 people have had the flu? according to the model, how many days until 700 people have had the flu? according to the model, how many days until 2000 people have had the flu? hint: one of your answers should be “never”

Explanation:

Step1: Find number of flu - infected after 20 days

Substitute $t = 20$ into $f(t)=\frac{1000}{1 + 999e^{-0.603t}}$.
$f(20)=\frac{1000}{1+999e^{-0.603\times20}}$
First, calculate the exponent: $- 0.603\times20=-12.06$. Then $e^{-12.06}\approx0$. So $f(20)=\frac{1000}{1 + 999\times0}=1000$.

Step2: Find time when 500 people have flu

Set $f(t)=500$, so $500=\frac{1000}{1 + 999e^{-0.603t}}$.
Cross - multiply: $500(1 + 999e^{-0.603t})=1000$.
Divide both sides by 500: $1 + 999e^{-0.603t}=2$.
Subtract 1 from both sides: $999e^{-0.603t}=1$.
Then $e^{-0.603t}=\frac{1}{999}$.
Take the natural logarithm of both sides: $-0.603t=\ln(\frac{1}{999})=-\ln(999)$.
Solve for $t$: $t=\frac{\ln(999)}{0.603}\approx\frac{6.907755}{0.603}\approx11.46\approx11$.

Answer:

After 20 days: 1000
Days until 500 people have flu: 11