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Thread: calculus 2 word problem

  1. #1
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    calculus 2 word problem

    Hi everyone,

    Can you please tell me if this is correct?

    A bacteria population starts with 400 bacteria and grows at a rate of r(t)=(450.268)e1.2567t bacteria per hour. How many bacteria will there be after three hours?

    I did:

    r(t)=ln450.268e1.2567(3)
    =(1.2567)(3)ln450.268
    =3.7701(6.10984261)

    r(t)=23.034 bacteria

    I'm not sure if I did it right. This seems too easy.

    Thank you
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  2. #2
    Super Member Rebesques's Avatar
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    This seems too easy.
    Yeap, I use the same criterion to see what's right


    Here you are given the rate of growth, and not the actual population. Call this N(t). The problem is of the form

    $\displaystyle N'(t)=ce^{\lambda t}$

    and sorry but I'm just too lazy to write down the constants So integrating this gives:

    $\displaystyle N(t)-N(0)=\frac{ce^{\lambda t}}{\lambda}$

    where N(0) is the initial population.
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  3. #3
    Grand Panjandrum
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    Quote Originally Posted by chocolatelover View Post
    Hi everyone,

    Can you please tell me if this is correct?

    A bacteria population starts with 400 bacteria and grows at a rate of r(t)=(450.268)e1.2567t bacteria per hour. How many bacteria will there be after three hours?

    I did:

    r(t)=ln450.268e1.2567(3)
    =(1.2567)(3)ln450.268
    =3.7701(6.10984261)

    r(t)=23.034 bacteria

    I'm not sure if I did it right. This seems too easy.

    Thank you
    Let the number of bacteria at time $\displaystyle t$ be $\displaystyle n(t)$, then the question tells you that:

    $\displaystyle
    \frac{dn}{dt}=450.268 ~e^{1.2567~t}
    $

    so:

    $\displaystyle
    n(t)=\frac{450.268}{1.2567} ~e^{1.2567~t}+C
    $

    and as at $\displaystyle t=0\ \ n=400$ we have:

    $\displaystyle
    C=400-\frac{450.268}{1.2567}
    $

    RonL
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  4. #4
    Grand Panjandrum
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    Quote Originally Posted by chocolatelover View Post
    Hi everyone,

    Can you please tell me if this is correct?

    A bacteria population starts with 400 bacteria and grows at a rate of r(t)=(450.268)e1.2567t bacteria per hour. How many bacteria will there be after three hours?

    I did:

    r(t)=ln450.268e1.2567(3)
    =(1.2567)(3)ln450.268
    =3.7701(6.10984261)

    r(t)=23.034 bacteria

    I'm not sure if I did it right. This seems too easy.

    Thank you
    As the rate of growth in numbers of bacteria is positive there must be
    more bacteria after 3 hours than you start with, so your answer cannot
    be right.

    RonL
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