The population of a certain species of insect is given by a differentiable function P, where P(t) is the number of insects in the population

Question

The population of a certain species of insect is given by a differentiable function P, where P(t) is the number of insects in the population, in millions, at time t, where t is measured in days. When the environmental conditions are right, the population increases with resect to time at a rate that is directly proportional to the population. Starting August 15, the conditions were favorable and the. population began increasing. On August 20, five days later, there were an estimated 10 million insects and the population was increasing at a rate of 2 million insects per day.

Required:
What is the differential equation that models this situation?

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Đan Thu 4 months 2021-07-17T08:49:48+00:00 1 Answers 0 views 0

Answers ( )

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    2021-07-17T08:50:57+00:00

    Answer:

    \frac{dP}{dt} \ =\  \frac{1}{5}P

    Step-by-step explanation:

    Given

    Variation: Directly Proportional

    Insects = 10 million when Rate = 2 million

    i.e. P = 10\ million when \frac{dP}{dt} = 2\ million

    Required

    Determine the differential equation for the scenario

    The variation can be represented as:

    \frac{dP}{dt} \ \alpha\ P

    Which means that the rate at which the insects increase with time is directly proportional to the number of insects

    Convert to equation

    \frac{dP}{dt} \ =\ k * P

    \frac{dP}{dt} \ =\ kP

    Substitute values for \frac{dP}{dt} and P

    2\ million = k * 10\ million

    Make k the subject

    k = \frac{2\ million}{10\ million}

    k = \frac{2}{10}

    k = \frac{1}{5}

    Substitute \frac{1}{5} for k in \frac{dP}{dt} \ =\ k * P

    \frac{dP}{dt} \ =\  \frac{1}{5}* P

    \frac{dP}{dt} \ =\  \frac{1}{5}P

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