# Measles pathogenesis curve by function f (see details for questions)?

## This question is relevant to Area Between Curves. I don't know what that constitutes under Socratic guidelines, so it is mentioned here. A patient infected with the measles virus who has some immunity to the virus has a pathogenesis curve that can be modeled by, for instance, $g \left(t\right) = 0.9 f \left(t\right) .$ If the threshold concentration of the virus required for infectiousness to begin is $\text{1210 cells/mL}$, on what day does this occur? Let ${P}_{3}$ be the point on the graph of $g$ where infectiousness begins. It has been shown that infectiousness ends at a point ${P}_{4}$ on the graph of $g$ where the line through ${P}_{3}$, ${P}_{4}$ has the slope $- 23$. On what day does infectiousness end? Compute the level of infectiousness for this patient.

Mar 27, 2017

Hi...this is only an attempt, completely without any (probable) connection with your question....because I didn't know the function $f \left(t\right)$ so I tried with something I found on the internet!!!

#### Explanation:

I found on the internet this:

Where $f \left(t\right)$is measured in number of infected cells per mL of blood plasma.

$g \left(t\right) = 0.9 \left[- t \left(t - 21\right) \left(t + 1\right)\right]$

I tried to plot this function:

Solving for $g \left(t\right) = 1210$ I got three solutions:
${t}_{1} = 11$ days
${t}_{2} = - 7$ days
${t}_{3} = 16$ days
I think it should be ${t}_{1} = 11$ days that should be the answer to question 1).

To answer question 2) we need a line passing through the above initial point:
$\left(11 , 1210\right)$ and slope $m - 23$.
We can use the general relationship from maths:
$y - {y}_{0} = m \left(x - {x}_{0}\right)$
where for us:
$g \left(t\right) - 1210 = - 23 \left(t - 11\right)$
$g \left(t\right) = - 23 t + 1463$

we can use this line to find the intercepts with:
$g \left(t\right) = 0.9 \left[- t \left(t - 21\right) \left(t + 1\right)\right]$ and get, substituting:
$- 23 t + 1463 = 0.9 \left[- t \left(t - 21\right) \left(t + 1\right)\right]$
with again three solutions:
${t}_{1} = 11$ days
${t}_{2} = - 8$ days
${t}_{3} = 17$ days
I would choose ${t}_{3} = 17$ days as answer for the end of infectiousness.

for question 3) I do not have any idea at all....sorry!