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Thread: Bacteria colony growth

  1. #1
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    Question Bacteria colony growth

    Hi,

    I hope someone can help. I'm trying to solve the following problem:

    Bacteria colony growth-screen-shot-2017-09-09-5.22.52-pm.png

    Apparently I have to create a function which would subtract K(t) from N(t), while taking into account that N(t) starts 60 minutes before K(t). Following that, I would need to use this function to find the critical values and determine the maximum value as usual. Found the steps for the solution here: https://answers.yahoo.com/question/i...6172646AAIFqMu

    The first half of this problem I don't understand as I thought that you could just solve the critical values for N(t) and then be able to find the maximum values as usual. I don't understand why you would need to subtract K(t) from N(t). Additionally, apparently N(t) also needs to have 60 added to it in order to take in account that it starts before K(t), but it makes more sense (in my head at least) for 60 to be added to K(t). Could someone please explain why this is the case?

    Please let me know!!
    - olivia
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  2. #2
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    Re: Bacteria colony growth

    It depends upon what "t" represents. If you take t to be the number of minutes after the drug is introduced then the time since the beginning of the experiment, 60 minutes earlier, is t+ 60: 2^{(t+ 60)/5}- e^{t/3}. If instead you take t to be the number of minutes since the beginning of the experiment, the time since the drug was introduced is t- 60: 2^{t/5}- e^{(t- 60)/3}.
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  3. #3
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    Re: Bacteria colony growth

    Alright cool, that makes sense. Thanks
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  4. #4
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    Re: Bacteria colony growth

    WAIT can someone please explain to me how 2^(t+60)/5-e^t/3 and (2^12)2^t/5 - e^t/3 are the same? I know how 2^(t+60)/5-e^t/3 was developed (pretty obvious based on the question) however I don't understand how one could realize that the second equation of (2^12)2^t/5 - e^t/3 would also work. The first term is different and I want to understand how one could manipulate this term while still resulting in the same value.
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