Question

Any kind souls willing to help me with a word problem?

Answer 1

Hello there

Answer 2

So, here it is:

Answer 3

You could save time by just posting the poblem statement and showing your own efforts.

Answer 4

The logistic growth function f(t) = 100571/1 + 5614 e^{-t}

describes the number of people, f(t), who have become ill with influenza t weeks after its initial outbreak in a particular community.

Answer 5

Perhaps so, but I've also wasted my fair share of time that way too ;)

Answer 6

Hence my use of different approach this time.

Answer 7

Fair enough...Is it \[\frac{100571}{1+5614e^{-t}}\]

Answer 8

Anyway...it seems to me that I should just plug in 3, which would result in 100571/1+5614e^(-3)

Answer 9

With the answer being, acc. to my calculations, 359 people.

Answer 10

Btw...should I put the denominator in parentheses?

Answer 11

When entering it into a calculator, that is

Answer 12

I believe that. Is there a disagreement? It could be in the definition of f(t) or in teh exact problem statement.

Answer 13

Well, I entered in a few other answers, which looked (and eventually turned out to be) wrong, probably because I didn't put the denom.. in parentheses. Now I've tried doing so, and come up with a more likely response. Just wanted to make sure though since we can only attempt each problem so many times.

Answer 14

But one more question...the next part of the question reads thusly:

What is the limiting size of the population that becomes ill?

Answer 15

I have no idea how to find the answer to this part.

Answer 16

That is a pretty flat logistic curve. It doesn't reach an infection rate of 1000 until t = 4 and then comes back down under 1000 by 14. Not reaqlly much going on there except the very middle.

The parentheses are VERY important. Order of Operations will save you!

Answer 17

Think about what happens as t increases without bound. That e^{-t} soon overcomes the neighboring 5614 and that whole term term goes away, leaving only 100571/1

Answer 18

Oh..

Answer 19

So...100571?

Answer 20

Yes. Limiting operations can be a bit tricky. Don't expect it to pop out. You must think it through.

Answer 21

But is it generally the numerator, in cases like this?

Answer 22

Like...are there any patterns one can apply when working with limiting operations?

Answer 23

Try it:

f(10) = 80114.22433...
f(20) = 100569.836274
f(30) = 100570.999947166
f(1000) = 100571.00000000

Answer 24

Oh...so it's just a question of plugging in especially large numbers

Answer 25

No, no. Don't try to make general rules. Youwill become confused. You must consider each case by itself. In THIS case, there was only one moving part, that e^(-t) in the denominator. Try this\[\frac{e^{t}}{1 + e^{t}}\]. What happens to that as t increases without bound?

Answer 26

Ummm...to the t? I'm honestly not sure.

Answer 27

Substitution can help, but that would be more useful as a demonstration that youare right, rather than a primary technique to find the limit.

Answer 28

In my example, after not very long, that e^t is WAY bigger than the one (1). Soon, the one (1) in the denominator is of virtually no significance and we are aproaching \[\frac{e^t}{e^t}\] and the limit is 1.

Answer 29

I see.

Answer 30

Well...thank you very much for your help, tkhunny! Havve a good day/night!!