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(1 point) A tank contains 23402340 L of pure water. Solution that contains 0.050.05 kg of sugar per liter enters the tank at the rate 55 L/min, 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)=y(0)= equation editorEquation Editor (kg) (b) Find the amount of sugar after t minutes. y(t)=y(t)= equation editorEquation Editor (kg) (c) As t becomes large, what value is y(t)y(t) approaching ? In other words, calculate the following limit. limt→[infinity]y(t)=limt→[infinity]y(t)= equation editorEquation Editor (kg)

User Margus
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1 Answer

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Let
S(t) be the amount of sugar in the tank at time
t.

a. The tank contains only pure water at the start, so
\boxed{S(0)=0}.

b. The inflow rate of sugar is


{S_(\rm in)}'=\left(0.05(\rm kg)/(\rm L)\right)\left(5(\rm L)/(\rm min)\right)=\frac1{40}(\rm kg)/(\rm min)

and the outflow rate is


{S_(\rm out)}'=\left(\frac S{2340}(\rm kg)/(\rm L)\right)\left(5(\rm L)/(\rm min)\right)=\frac S{468}(\rm kg)/(\rm min)

so the net rate at which
S(t) changes over time is governed by


S'=\frac1{40}-\frac S{468}\implies S'+\frac S{468}=\frac1{40}

Multiply both sides by
e^(t/468),


e^(t/468)S'+(e^(t/468))/(468)S=(e^(t/468))/(40)

and condense the left side as the derivative of a product,


\left(e^(t/468)S\right)'=(e^(t/468))/(40)

Integrate both sides to get


e^(t/468)S=(117e^(t/468))/(10)+C

and solve for
S:


S=(117)/(10)+Ce^(-t/468)

With
S(0)=0, we find
C=-(117)/(10)=-11.7, so that


\boxed{S(t)=11.7-11.7e^(-t/468)}

c. As
t\to\infty, the exponential term will converge to 0, leaving a fixed amount of 11.7 kg of sugar in the solution.

User Jeff Schumacher
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7.7k points
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