Question

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.

Answer 1

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.