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Numerade Educator



Problem 51 Hard Difficulty

In Example 5, we modeled a measles pathogenesis curve by a function $ f $. 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(t) = 0.9 f(t) $.
(a) If the same threshold concentration of the virus is required for infectiousness to begin as in Example 5, on what day does this occur?
(b) Let $ P_3 $ be the point of the graph of $ g $ where infectiousness begin. 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 same slope as the line through $ P_1 $, $ P_2 $ in Example 5(b). On what day does infectiousness end?
(c) Compute the level of infectiousness for this patient.


a. 11th day (11.250)
b. $-23(x-11)$
c. 4387.2


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Video Transcript

Okay, so, uh, this problem gives us that GFT for someone has already been infected. The infectious. Ah, the path of Genesis Kerr would just be a 90% of the curve given an example. Five. Um so how it it said in part A it the same threshold. Concentration is required. So that threshold concentration was 12. 10. So we need to figure out when that threshold is reached. And so if we plot 0.9 times FFT And so I did negative zero point 90 times, t minus 21 times t plus one. Okay. And so in doing that, I found that my t value and that gives me 12 10. Now, this craft, each side of that equation, uh, my my ex, my T value would have Teoh equal 11.258 So it would happen somewhere on the 11th day. The infection sickness would start. Can somebody is, um, just for sake of keeping track 11 point to find me? That's that's party as opposed to the 10th day. An example. Five. Um, Barbie says p three is that point where the infectiousness began. So be I'm gonna dio if you look back to our chart. Um, it says infectiousness began on day between day 10 and 11. So they put in 10. So we're just gonna plug in 11 into this function and said my P three is gonna be 11 and then the actual number that would give it for 11. Um, we would give us 11. 88 concentration. So I'm gonna use that point for P three. Um and so we're looking for a similar slope to the slope. An example? Five. Let's go back to example. Five. That slope waas. Um, negative. 23. And so the equation of that line is just gonna be a y minus 11. 88 equals negative 23 times T, but money's X in this case minus 11. Okay, so we comply. We can plot that into a graphing calculator and see where it intersex. Um, so let's find out where that would be. So if we graph that on the same graph, Okay, we're gonna have a p for around days. 17. So be 17.377 Um, and so I'm just gonna go ahead and say my p for is gonna be at 17 for my ex Dykes. It'll be happening on day 17. And if I plugged that into my, um, original equation, that's gonna give me 1101.6, because now we have a P three and p four going from day 11 today. 17 Barbie, What date is the infectious in this? And it would end on the 17th day and part c compute the level of infectiousness, if you remember from example five. The level of infectiousness is the area between the curve and that line. So we have the equation for the line. So I'm just gonna go back in, figure out issues, the original equation instead of this point for p four. Okay, so we're gonna take the integral of GFT, which remember, is 90% of fft and we're gonna subtract our equation here of the line. We're gonna integrate that from 11 to 17. So that's gonna be minus negative 23 x the negative 23 times. Negative 11 And then we're going Teoh 2 53 and they were going to add 11. 88 to that. Be 14 41. Okay, that'll give us our level of infectiousness. So Let's just do that side. A lot of side work that goes into this. So, um, we're gonna have negative 0.9 times. They use the information from example. Three eso We have negative R t cubed. Ah, minus 20 squared. Um, minus two t. So I'm kind of doing this on the five using the example. Um, and they were gonna be adding 23 x. It's attracting 14. 41. Okay, so that's gonna give us negative 0.9 t. Cute. Uh, plus 18 t squared. Those 1.80 Actually, this is gonna be positive, Teoh, from our original equation. Okay, I'm gonna I feel like I'm losing some stuff, so I'm gonna go back and reevaluate what GFT is. Um, so I apologize for this. I just wanna make sure way get it right. Not like I'm losing some negatives and losing some numbers here. Eso GFT. If we go back up to the top, let's just get that done. Ah is equal to negative 0.90 times T minus 21 times T plus one, which is no 0.90 times T squared, minus 21 t plus T minus 21. Okay, that's gonna be negative. 0.9 t times t squared finest 20 T minus 21 which in turn gives us negative 0.9 t cubed plus 18. T square, 0.9 comes 21 is 18.9 was 18.90. Okay, there's fungi of tea. And they were going Teoh ad 23 t and take away 14. 41. Okay, so our final equation that were integrating from 11 to 17 is negative. 0.9 t caved plus 18 t squared. 23 plus 18.9. It's 41.9 minus 14. 41. DT, This is a polynomial. Also to integrate it. We're just gonna dio kind of the powerful. She just adding one teach exponents and dividing by that number. So we're gonna have Can I get a 0.9? It's about about four times teacher before, plus 60 cubed. That's just 18. About about three was 41.9. About about Teoh a T squared minus 14. 21 t. We're evaluating that from 11 to 17. Okay, so evaluating at 17 we're gonna get a value of negative 17 2.125 Okay, we're gonna subtract what we get. 11 also be negative. Most of the negative 6089.3 to 5. Is that level of infectiousness. It's just attractors. So we have negative. 17 2.25. The 6089.3 to 5. The level of infectiousness is gonna be 4387 0.2. That's gonna be cells per milliner days today.