Logistic Growth model question. Please help!?
on a college campus of 5000 students, one student returns from vacation with contagious flu virus. The speed of the virus is modeled by:
y=5000/(1+4999e^(-0.8*t)) , 0 is smaller or equal to t
where y is the total number infected after t days. The college will cancel classes when 40% or more of the students are ill.
a. How many students are infected after 5 days?
b. After how many days will the college cancel classes?
(verifying using your calculator)
I don't know how to get the right answer :(
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Answers & Comments
Verified answer
a) No thinking needed for this part. All you have to do is write the number 5 where t was.
That is called substituting, which is a really easy thing to do. It then looks like this.
y = 5000/[1+4999e^(-0.8*5)]
Then simplify -0.8*5 as -4
y = 5000/[1+4999e^(-4)]
y represents the total number of students infected after t days
Just let a calculator work out what y is.
y = 54.01 so that means 54 since there must be a whole number of students.
b) The college will cancel classes when 40% or more of the students are ill.
There are 5000 students and 40% of that is 2000.
Our job now is to find t when y = 2000, so start by substituting that value for y
2000 = 5000/[1+4999e^(-0.8*t)]
Rearrange things to work towards finding t
1+4999e^(-0.8*t) = 5000/2000 = 5/2
4999e^(-0.8*t) = 4999/e^(0.8*t) = 3/2
e^(0.8*t) = (2/3)4999 = 9998/3
Take logs to base e of both sides
(0.8*t) = (4/5)t = ln[9998/3]
t = (5/4)ln[9998/3]
Now use that calculator
t ~ 10.13 days
With t = 10 days students ill = 1868
With t = 11 days students ill = 2851
Although the 2000 mark is only exceeded with t = 11 days, it is noticeable that nearly a thousand extra get ill for that 1 last day. A wise headmaster would cancel classes at a much lower percentage.
Regards - Ian H