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Math Help - 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

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

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

    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 t be n(t), then the question tells you that:

    <br />
\frac{dn}{dt}=450.268 ~e^{1.2567~t}<br />

    so:

    <br />
n(t)=\frac{450.268}{1.2567} ~e^{1.2567~t}+C<br />

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

    <br />
C=400-\frac{450.268}{1.2567}<br />

    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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