PRE-CALCULUS Transmission rates

Question

4.)Explore the consequences of different transmission rates. Start with 10 people initially

infected, and compare a 1%, 2.5%, and 5% transmission rate (you may wish to use Desmos or a

similar graphing utility to look at the graphs of these functions and see the growth, and you are

welcome to include your graphs for visual confirmation, but you must support your answers to

(a)-(b) below algebraically):

a) How long until 100 people are infected?

b) How long until 1000 people are infected?

5.)Find the doubling time for each of the three scenarios above (1%, 2.5%, and 5% transmission

rate).

Answer 1

Ok, let us try and understand how this question works conceptually. I'll leave you to fill in the details.

Let us say that we start off with N₀ infected individuals.

Since viruses transmit (thereby infecting more people over time), the central problem we are dealing with is to estimate the number of infected people after a given time has elapsed since the initial discovery of infection in N₀ people.

Note: 'time elapsed' will measured in multiples of some basic unit of time. This could be days, weeks,.. We'll call the unit time Δt.

The 'rate of transmission' is given as r. This means that if we know N(t), the infected population at some point in time t, then one unit time later, the estimated infected population will be N(t + Δt) = N(t) (1 + r)

This in turn implies that the estimated infected population some T units of time after t is simply given as :

N(t + T·Δt) = N(t) (1+r)^T.

Finally, we wish to know how much time will elapse before at least N people will be infected, starting with an initial infected population of N₀ (where N > N₀).

That is, we wish to find T such that N(t₀ + T·Δt) ≥ N, where N(t₀) = N₀

Or : N₀ (1+r)^T ≥ N ⇒ (1 + r)^T ≥ N / N₀.

Now, if A ≥ B , and both are positive, then we can take the logarithm of both sides, and the inequality should still hold : ln (A) ≥ ln (B).

So, the above inequality reduces to T ln(1 + r) ≥ ln(N / N₀)

Since 1 + r > 1, ln(1 + r) ≥ 0. So we may divide both sides by ln(1 + r) without affecting the direction of the inequality. So finally, we obtain :

T ≥ ln(N / N₀) / ln(1 + r).

That is, T is the lowest integer greater than or equal to the RHS (which will in general be a real number, not necessarily an integer).

Hope this helps.