A tank contains 100 grams of a substance dissolved in a large amount of water. T

Question

A tank contains 100 grams of a substance dissolved in a large amount of water. The tank is filtered in such a way that water drains from the tank, leaving the substance behind in the tank. Consider the volume of the dissolved substance to be negligible. At what rate is the concentration (grams/liter) of the substance changing with respect to time in each scenario? (a) the rate after 5 hours, if the tank contains 50 L of water initially, and drains at a constant rate of 4 L/hr? (b) the rate at the instant when 20 liters remain, if the water is draining at 2.4 L/hr at that instant (c) the rate in scenario (b), if the unknown substance is also being added at a rate of 40 g/hr (and there are 100 grams in the tank at that instant) Find C, the concentration, as a function of V, the volume of the tank. Translating the given facts to mathematical symbols in each question part, what type of differentiation is needed to find dC/dt?

Explanation / Answer

(a)
Let K = concentration
K = 100 / V

. . . V = 50 - 4t

K = 100 / (50 - 4t)

dK/dt = 100 / (25-2 t)^2

. . . t = 5

dK/dt = 100 / (25 - 2*5)^2 = 4/9 grams per liter per hour

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(b)

K = 100 / (20 - 2.4 t)

dK/dt = 240 / (20 - 2.4 t)^2

. . . t = 0

dK/dt = 240 / (20 - 0)^2 = 3/5 grams per liter per hour

************
c)

K = (100 + 40t) / (20 - 2.4t)

dK/dt = 1040/ (20 - 2.4 t)^2

. . . t = 0

dK/dt = 1040 / (20 - 0)^2 = 13/5 grams per liter per hour

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