## soccergal12 Group Title Peter's chickens got the bird flu and he is rying to stop the epidemic. He knows that the function f(t) = 300 / ( 10 + 20e^(-1.5t) ) describes the number of his chickens who are sick t weeks after the initial outbreak. (a) How many chickens became sick when the flu epidemic has began? I got 10/7 for the answer. does that seem right??? (b) In how many weeks will 100 chickens be sick? (c) What is the maximum number of chickens that will become ill? 2 years ago 2 years ago

1. rafabc02 Group Title

For part (a) we just have to plug in t=0, so: 300/(10+20)=300/30=10 Part (b): Plut t=100 and solve. Part(c): Take the derivative and set it equal to zero.

2. rafabc02 Group Title

@soccergal12 Did that help?

3. soccergal12 Group Title

yes it did, thank you

4. rafabc02 Group Title

5. soccergal12 Group Title

im just unsure about the derivative; i'm not sure how to do it entirely

6. soccergal12 Group Title

and in question b, aren't i finding t, not f(100) ?

7. soccergal12 Group Title

scratch that last question .. i'm just unsure about how to do the derivative : where do i begin?

8. soccergal12 Group Title

i got ( 9000e^-1.5t ) / (10 + 20e^-1.5t )^2

9. rafabc02 Group Title

You are right, about part (b), sorry about that, we have to do this: 100=300 / ( 10 + 20e^(-1.5t) ) And solve for t Part (c): $f(t) = 300( 10 + 20e^{-1.5t} )^-1$ so the derivative is: $300*(-1)*(10 + 20e^{-1.5t} )^{-2}*20e^{-1.5t}(-1.5)$

10. rafabc02 Group Title

Which is the same as: $\frac{9000e^{-1.5t}}{(10+20e^{1.5t})^2}$

11. rafabc02 Group Title

So you did it great! Any doubt?

12. soccergal12 Group Title

okay thanks! sorry i have another question about part b : to find t, i got down to e^-1.5t = -7/20 ; could you do this question and let me know if you get the same answer? also, where do i go from here: do i ln both sides?

13. rafabc02 Group Title

$100=\frac{300}{ ( 10 + 20e^{-1.5t} )}\rightarrow10 + 20e^{-1.5t} =\frac{300}{100}$ So then: $20e^{-1.5t}=20$ $e^{-1.5t}=1$

14. soccergal12 Group Title

doesn't it equal 20^e −1.5t = -7 ?

15. rafabc02 Group Title

We start from here: $100=\frac{300}{10+20e^{-1.5t}}$ Then we have this: $10+20e^{-1.5t}=\frac{300}{100}$ So then, $20e^{-1.5t}=3-10=-7$ You are right! Sorry about my confusion! But then, how do we solve it?

16. rafabc02 Group Title

I think that this function never reaches 100. Take a look here: http://www.wolframalpha.com/input/?i=100+%3D+300+%2F+%28+10+%2B+20e%5E%28-1.5t%29+%29

17. soccergal12 Group Title

so would i state that the function doesn't ever reach 100?

18. rafabc02 Group Title

I think so. You can see it on the graph. Any other doubt?

19. Jemma6 Group Title

I have this same question, but it has been changed to: f(t) = 300 / ( 15 - 20e^(-0.07t) ) When I have this information though, the answer I get for part A is negative. That doesn't make sense. Can anyone help me figure out how to solve this with the changes made?

20. soccergal12 Group Title

I get a negative answer too, and my function has changed too. could anyone help us figure this out ?

21. dumbcow Group Title

this looks like a logistic function it approaches an upper limit as t goes to infinity the term "20e^(-1.5t)" goes to 0 leaving 300/10 = 30 the maximum possible sick chickens is 30 it will never be 100 so part b) is never