Section 2.3 Modeling with First order: Problem 3 Previous Problem List Next (1 p

Question

Section 2.3 Modeling with First order: Problem 3 Previous Problem List Next (1 point) A tank contains 2420 L of pure water. Solution that contains 0.09 kg of sugar per liter enters the tank at the rate 7umin, and is thoroughly mixed into it. The new solution drains out of the tank at the same rate. (a) How much sugar is in the tank at the begining? y(0) (kg) (b) Find the amount of sugar after t minutes. y(t) (kg) (c) As t becomes large, what value is y(t) approaching In other words, calculate the following limit. lim y(t) (kg) Note: can eam partial credit on this problem

Explanation / Answer

3) If r is amount of sugar.

dr/dt = amount of sugar going in - amount of sugar going out.

Amount of sugar going in = 0.09 kg/L * 7L/min = .63 kg/min

Amount of sugar going out depends on concentration.

Concentration is r/2420 in kg/L.

But there are 7 L/min going out.

Amount of sugar going out = 7r/2420.

dr/dt = 0.63 - 7r/2420 = (1524.6 - 7r)/2420

Separating variables, dr/(1524.6 - 7r) = dt/2420

=> -1/7*ln(1524.6 - 7r) = t/2420 + c

=> ln(1524.6 - 7r) = -7t/2420 + d       [absorb the negative sign into the constant]

c and d are constants

=> 1524.6 - 7r = e(-7t/2420 + d)

=> r = 1524.6/7 - 1/7*ede(-7t/2420)

=> r = 217.8 + a.e(-7t/2420) where a is constant

a) At t = 0, r = 0

so a = -217.8

b) r = 217.8(1 - e(-7t/2420) )

c) As t , r = 217.8 kg

4) If r is amount of salt.

dr/dt = amount of salt going in - amount of salt going out.

Amount of salt going in = 0.025 kg/L * 6L/min = .15 kg/min

Amount of salt going out depends on concentration.

Concentration is r/1000 in kg/L.

But there are 6 L/min going out.

Amount of salt going out = 6r/1000.

dr/dt = 0.15 - 6r/1000 = (150 - 6r)/1000

Separating variables, dr/(150 - 6r) = dt/1000

=> -1/6*ln(150 - 6r) = t/1000 + c

=> ln((150 - 6r)) = -6t/1000 + d       [absorb the negative sign into the constant]

c and d are constants

=> 150 - 6r = e(-6t/1000 + d)

=> r = 150/6 - 1/6*ede(-6t/1000)

=> r = 25 + a.e(-6t/1000) where a is constant

a) At t = 0, r = 50

=> 50 = 25 + a

=> a = 25

b) r = 25(1 + e-0.006t)

c) As t , r = 25 kg

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