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March 28, 2017

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A population of 500 E. coli bacteria doubles every 15 minutes. Use this information to find an expression for this population growth. Using this expression, find what the population would be in 87 minutes. Use an exponential model.


so we're supposed to use P(t)= (Po)(e)^kt

so P(o) would be 500,
and then would t be 15 or 0.25 and wowhat would k be? 2?

  • Calculus - ,

    in P(t) = 500 e^kt
    it depends whether you want to to be in minutes or hours.

    let's do it for minutes, then
    1000 = 500 e^15k
    2 = e^15k
    15k = ln2
    k = (ln2)/15

    if you want t to be hours then use .25 for t in
    1000 = 500 e^.25K
    .
    .
    K = ln2/.25 or K = 4ln2

  • Calculus - ,

    okay, so then in 87 minutes what would it be?

  • Calculus - ,

    let's use our first one, the one in minutes
    P(87) = 500 e^(87*ln2/15)
    = 500 e^4.020254
    = 27857.6

    is our answer reasonable??

    remember it doubles every 15 min

    so after 15 min --> 1000
    after 30 min ---> 2000
    after 45 min ---> 4000
    after 60 min ---> 8000
    after 75 min ---> 16000
    after 90 min ---> 32000

    we could check to see if we get 32000 for 90 minutes

    P(90) = 500 e^(90*ln2/15)
    = 500(64) = 32000 YES!!!!

  • Calculus - ,

    whao! that makes so much sense.
    just a quick question, how do u know what e is?

    thanks so much
    <3333333333

  • Calculus - ,

    e is one of these very strange transcendental numbers like pi

    one defn of e is
    1 + 1/1! + 1/2! + 1/3! + ... to infinitity

    where something like 5! = 1*2*3*4*5

    or
    e = Limit (1 + 1/n)^n where n approaches infinitity

    The Swiss/German mathematician Euler spent a lot of time with the concept of that number and e is often called Euler's number

  • Calculus - ,

    e is approximately 2.71828

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