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soccergal12

  • 3 years ago

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?

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  1. rafabc02
    • 3 years ago
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    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
    • 3 years ago
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    @soccergal12 Did that help?

  3. soccergal12
    • 3 years ago
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    yes it did, thank you

  4. rafabc02
    • 3 years ago
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    OK, awsome! Glad to help you!

  5. soccergal12
    • 3 years ago
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    im just unsure about the derivative; i'm not sure how to do it entirely

  6. soccergal12
    • 3 years ago
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    and in question b, aren't i finding t, not f(100) ?

  7. soccergal12
    • 3 years ago
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    scratch that last question .. i'm just unsure about how to do the derivative : where do i begin?

  8. soccergal12
    • 3 years ago
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    i got ( 9000e^-1.5t ) / (10 + 20e^-1.5t )^2

  9. rafabc02
    • 3 years ago
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    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
    • 3 years ago
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    Which is the same as: \[\frac{9000e^{-1.5t}}{(10+20e^{1.5t})^2}\]

  11. rafabc02
    • 3 years ago
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    So you did it great! Any doubt?

  12. soccergal12
    • 3 years ago
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    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
    • 3 years ago
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    \[100=\frac{300}{ ( 10 + 20e^{-1.5t} )}\rightarrow10 + 20e^{-1.5t} =\frac{300}{100}\] So then: \[20e^{-1.5t}=20\] \[e^{-1.5t}=1\]

  14. soccergal12
    • 3 years ago
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    doesn't it equal 20^e −1.5t = -7 ?

  15. rafabc02
    • 3 years ago
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    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
    • 3 years ago
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    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
    • 3 years ago
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    so would i state that the function doesn't ever reach 100?

  18. rafabc02
    • 3 years ago
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    I think so. You can see it on the graph. Any other doubt?

  19. Jemma6
    • 3 years ago
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    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
    • 3 years ago
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    I get a negative answer too, and my function has changed too. could anyone help us figure this out ?

  21. dumbcow
    • 3 years ago
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    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

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