Question

During optimal conditions, the rate of change of the population of a certain organism is proportional to the population at time t, in hours. At time t = 0 hours, the population is 300. At time t = 24 hours, the population is 1000. At what time t is the population 500?
(A) t = 2112
(B) t = 24 In 500 In 1000
(D) t = 300e (9)

Answers

  1. Answer:

    The population is of 500 after 10.22 hours.

    Step-by-step explanation:

    The rate of change of the population of a certain organism is proportional to the population at time t, in hours.

    This means that the population can be modeled by the following differential equation:

    \frac{dP}{dt} = Pr

    In which r is the growth rate.

    Solving by separation of variables, then integrating both sides, we have that:

    \frac{dP}{P} = r dt

    \int \frac{dP}{P} = \int r dt

    \ln{P} = rt + K

    Applying the exponential to both sides:

    P(t) = Ke^{rt}

    In which K is the initial population.

    At time t = 0 hours, the population is 300.

    This means that K = 300. So

    P(t) = 300e^{rt}

    At time t = 24 hours, the population is 1000.

    This means that P(24) = 1000. We use this to find the growth rate. So

    P(t) = 300e^{rt}

    1000 = 300e^{24r}

    e^{24r} = \frac{1000}{300}

    e^{24r} = \frac{10}{3}

    \ln{e^{24r}} = \ln{\frac{10}{3}}

    24r = \ln{\frac{10}{3}}

    r = \frac{\ln{\frac{10}{3}}}{24}

    r = 0.05

    So

    P(t) = 300e^{0.05t}

    At what time t is the population 500?

    This is t for which P(t) = 500. So

    P(t) = 300e^{0.05t}

    500 = 300e^{0.05t}

    e^{0.05t} = \frac{500}{300}

    e^{0.05t} = \frac{5}{3}

    \ln{e^{0.05t}} = \ln{\frac{5}{3}}

    0.05t = \ln{\frac{5}{3}}

    t = \frac{\ln{\frac{5}{3}}}{0.05}

    t = 10.22

    The population is of 500 after 10.22 hours.

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