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Math Help - Exponential function

  1. #1
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    Exponential function

    The spread of head lice at a particular school was traced back to a sports trip to a neighbouring school. The spread amongst the students at the school could be modelled by the exponential functionL(t) = 4e^(8t - t^2)k (^ - square) , where t represents the number of weeks after the sports trip, and k is the rate of growth

    How many students were initially infected? Justify your answer.
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  2. #2
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    Quote Originally Posted by Snowboarder View Post
    The spread of head lice at a particular school was traced back to a sports trip to a neighbouring school. The spread amongst the students at the school could be modelled by the exponential functionL(t) = 4e^(8t - t^2)k (^ - square) , where t represents the number of weeks after the sports trip, and k is the rate of growth

    How many students were initially infected? Justify your answer.
    In the equation y=Ce^{kt} ...C is always the initial value...since when t=0 y=Ce^{0k}=C
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  3. #3
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    oki here is another one:

    After 2 weeks, it was found that 56 students had become infected.
    Calculate the rate of growth, k (to 4 dp).
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  4. #4
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    Quote Originally Posted by Snowboarder View Post
    oki here is another one:

    After 2 weeks, it was found that 56 students had become infected.
    Calculate the rate of growth, k (to 4 dp).
    1. Do yourself and do us a favour: If you have a new problem please start a new thread. Otherwise no member of the forum can see that you need some help.

    2. Plug in t = 2 into the given equation of the function:

    L(t) = 4e^{(8t-t^2) \cdot k}~\buildrel {t = 2} \over \longrightarrow~ L(2)=56= 4e^{(16-4) \cdot k}

    56 = 4e^{12k}~\iff~14=e^{12k}~\iff~\ln(14)=12k ... \iff~ \boxed{k=\frac1{12} \ln(14) \approx 0.21992 \approx 22\%}
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  5. #5
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    thank you earboth .

    Does anybody know when the outbreak was at its worst?
    When the outbreak was effectively over. Include any assumptions.
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  6. #6
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    Quote Originally Posted by Snowboarder View Post
    thank you earboth .

    Does anybody know when the outbreak was at its worst?
    1. Calculate the first derivative. Since you have to use the chainrule one of the factor is (8-2t). This is the only factor which can become zero.
    Therefore you'll get t = 4 if you calculate L'(t) = 0.

    When the outbreak was effectively over. Include any assumptions.
    You have first to define a state called "outbreak is over". Assume the outbreak has passed if only one person is actually infected:

    1 = 4e^{(8t-t^2) \cdot \frac{\ln(14)}{12}}

    I haven't the time yet to do the necessary calculatations in detail but you get t \approx -0.72267~ or~ t\approx 8.72267

    The first value (measured in weeks!) gives the start of the infection and the second value marks the actual end of the infection.
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