Problem Statement
The spread of a viral rumor in an online community of 5000 users is modeled by the logistic function:
$$N(t) = \frac{5000}{1 + 99 e^{-0.3 t}}$$
where $N(t)$ is the number of users who have heard the rumor after $t$ hours.
**(a)** [2 marks] Find the number of users who initially started the rumor at $t = 0$.
**(b)** [3 marks] Find the time $t$ when half of the community has heard the rumor.
**(c)** [3 marks] Find the maximum rate at which the rumor is spreading (in users per hour), and state the time at which this peak rate occurs.
Verified Solution & Marking Scheme
Part (a) [M1]
Evaluate at $t = 0$: $N(0) = \frac{5000}{1 + 99(1)} = \frac{5000}{100}$
Part (a) [A1]
$N(0) = 50\text{ users}$
Part (b) [M1]
Set $N(t) = 2500$: $\frac{5000}{1 + 99 e^{-0.3 t}} = 2500 \implies 1 + 99 e^{-0.3 t} = 2$
Part (b) [M1]
$99 e^{-0.3 t} = 1 \implies e^{-0.3 t} = \frac{1}{99}$
Part (b) [A1]
$-0.3 t = -\ln 99 \implies t = \frac{\ln 99}{0.3} \approx \frac{4.595}{0.3} \approx 15.3\text{ hours}$
Part (c) [R1]
In a logistic model $\frac{L}{1 + C e^{-kt}}$, maximum growth rate occurs at the inflection point where $N = \frac{L}{2} = 2500$, which occurs at $t = 15.3\text{ hours}$.
Part (c) [M1]
Differential equation form: $\frac{dN}{dt} = k N \left(1 - \frac{N}{L}\right) = 0.3(2500)\left(1 - \frac{2500}{5000}\right) = 750(0.5)$
Part (c) [A1]
Maximum growth rate $= 375\text{ users/hour}$ occurring at $t = 15.3\text{ hours}$