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Math Help - Integral

  1. #1
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    Integral

    A bacteria population starts with 400 bacteria and grows at a rate of
    r(t)= (450.268) e^1.12567t bacteria per hour.
    How many bacteria will there be after 3 hours?
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  2. #2
    Senior Member ecMathGeek's Avatar
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    Quote Originally Posted by camherokid View Post
    A bacteria population starts with 400 bacteria and grows at a rate of
    r(t)= (450.268) e^1.12567t bacteria per hour.
    How many bacteria will there be after 3 hours?
    For this, you need only to plug in 3 for t:

    r(3) = (450.268) e^(1.12567*3) = 13185.2355160

    Round this number to as many digits as you need for you solution.
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  3. #3
    Senior Member ecMathGeek's Avatar
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    Quote Originally Posted by camherokid View Post
    A bacteria population starts with 400 bacteria and grows at a rate of
    r(t)= (450.268) e^1.12567t bacteria per hour.
    How many bacteria will there be after 3 hours?
    How can the rate of growth be
    r(t) = (450.268) e^(1.12567t)

    The coefficient (450.268) of the exponential function (e^(1.12567t)) should be equal to the initial population (400).
    I don't see why you would have 450.268 in front of the e.
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  4. #4
    Senior Member ecMathGeek's Avatar
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    Quote Originally Posted by ecMathGeek View Post
    For this, you need only to plug in 3 for t:

    r(3) = (450.268) e^(1.12567*3) = 13185.2355160

    Round this number to as many digits as you need for you solution.
    Oops... I just noticed this is the growth rate and not the population at time t.

    r(t) = (450.268) e^(1.12567*t)

    P(t) = INT r(t) dt = INT (450.268) e^(1.12567*t) dt
    P(t) = (450.268/1.12567)e^(1.12567*t) + C

    At t = 0, P(t) = 400
    400 = (400)e^(1.12567*0) + C --> C = 0

    P(t) = (400)e^(1.12567*t)

    Solving for the population at t = 3, we get
    P(3) = (400)e^(1.12567*3) = 11713.2334663
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