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26 December, 03:16

The rate at which a duck gains weight is proportional to the difference between its adult weight and its current weight. At time t=0, when the duck is first weighed, it's weight is 20 grams. If B (t) is the weight of the duck, in grams, at time t days after it is first weighed, then "dB/dt = 1/5 (100-B) ". Let y=B (t) be the solution to the differential equation above with the initial condition B (0) = 20.

Is the duck gaining weight faster when it weighs 40 grams or when it weighs 70 grams? Explain your reasoning.

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  1. 26 December, 03:39
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    DB/dt = 1/5 (100 - B)

    5 / (100 - B) dB = dt

    -5 ln (100 - B) = t + C

    Since B (0) = 20, then

    -5 ln (100 - 20) = C

    i. e. C = - 5 ln 80

    Thus - 5 ln (100 - B) = t - 5 ln 80

    or t = 5 ln 80 - 5 ln (100 - B) = 5 ln (80 / (100 - B))

    When the weight = 40 grams

    t = 5 In (80 / 100 - 40) = 5 In (80 / 60) = 5 ln (4/3) = 1.438 days

    Rate of weight gain = 40 / 1.438 = 27.8 grams per day

    When the weight = 70 grams

    t = 5 ln (80 / 100 - 70) = 5 ln (80 / 30) = 5 ln (8/3) = 4.9 days

    Rate of weight gain = 70/4.9 = 14.27 grams per day.

    Therefore, the duck gains weight faster when it weighs 40 grams.
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