The population of a community is known to increase at a rate proportional to the number of people present at time t. The initial population

Question

The population of a community is known to increase at a rate proportional to the number of people present at time t. The initial population P0 has doubled in 5 years. Suppose it is known that the population is 9,000 after 3 years. What was the initial population P0? (Round your answer to one decimal place.) P0 = 5937.8 What will be the population in 10 years? (Round your answer to the nearest person.) 23751 persons How fast is the population growing at t = 10? (Round your answer to the nearest person.) 3293 persons/year

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Allison 2 months 2021-10-13T09:24:22+00:00 1 Answer 0 views 0

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    2021-10-13T09:25:37+00:00

    Answer:

    P0 = 5937.8 people

    P(t = 10) = 23751 people

    P'(t = 10) = 3293 persons/year

    Step-by-step explanation:

    Let the population has the formula of

    P = P_0e^{kt}

    Where P0 is the initial population at t = 0 and k is the constant that we are looking fore.

    Since the population doubled after t = 5 years

    P = 2P_0

    2P_0 = P_0e^{k5}

    e^{5k} = 2

    k = ln2/5 = 0.1386

    So after t = 3 years, population is P = 9000:

    9000 = P_0e^{0.1386*3}

    P_0 = \frac{9000}{e^{0.1386*3}} = 5937.8

    After 10 years, population would be quadtripled (10 years is 2 times of 5 years):

    P(10) = 4P_0 = 5937.8*4 = 23751

    The rate of change in population is the derivative of the population function with respect to t

    P'(t) = P_0ke^{kt} = 5937.8*0.1386e^{0.1386t} = 823.15 e^{0.1386t}

    So after t = 10 years the rate of change in population would be

    P'(10) = 823.15 e^{0.1386*10} = 3293 persons/years

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