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6 January, 05:22

The population of bacteria in a culture grows at a rate proportional to the number of bacteria present at time t. After 1 day, it is observed that 200 bacteria are present. After 3 days, 2000 bacteria are present. What was the initial number of bacteria?

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  1. 6 January, 05:28
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    The initial number of bacteria is 63.

    Step-by-step explanation:

    The population of the bacteria can be modeled by this following differential equation:

    dP/dt = Pr

    where r is the growth rate.

    Solving this equation by the method of variable separation, we end up with:

    dP/P = rdt

    Integrating both sides, we have equation 1)

    1) ln P = rt + P0

    Where P0 is the initial number of bacteria

    We need to isolate P in equation 1), so we do this

    e^ (ln P) = e^ (rt + P0)

    P (t) = P0 (e^ (rt))

    Given that P (1) = 200

    P0e^ (r) = 200

    The same for P (3) = 2000

    P0e^ (3r) = 2000

    Now we solve the following system of two equations, where we want the find two values (P0 and r)

    1) P0e^ (r) = 200

    2) P0e^ (3r) = 2000

    Isolating P0 in 1) in 1 and replacing it in 2), we have

    P0 = 200e^ (-r)

    200e^ (-r) e^ (3r) = 2000

    We know that e (a) e (b) = e (a+b), so e^ (-r) e^ (3r) = e (2r), so

    e (2r) = 10

    we need to find r, so we put ln in both sides of the equation

    ln (e (2r)) = ln (10)

    2r = 2.30

    r = 1.15

    Now, from 1), we know that

    P0 = 200e^ (-r) = 200e^ (-1.15) = 63

    So the initial number of bacteria is 63.
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