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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)

Respuesta :

LammettHash

Let [tex]S(t)[/tex] be the amount of sugar in the tank at time [tex]t[/tex].

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

b. The inflow rate of sugar is

[tex]{S_{\rm in}}'=\left(0.05\dfrac{\rm kg}{\rm L}\right)\left(5\dfrac{\rm L}{\rm min}\right)=\dfrac1{40}\dfrac{\rm kg}{\rm min}[/tex]

and the outflow rate is

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

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

[tex]S'=\dfrac1{40}-\dfrac S{468}\implies S'+\dfrac S{468}=\dfrac1{40}[/tex]

Multiply both sides by [tex]e^{t/468}[/tex],

[tex]e^{t/468}S'+\dfrac{e^{t/468}}{468}S=\dfrac{e^{t/468}}{40}[/tex]

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

[tex]\left(e^{t/468}S\right)'=\dfrac{e^{t/468}}{40}[/tex]

Integrate both sides to get

[tex]e^{t/468}S=\dfrac{117e^{t/468}}{10}+C[/tex]

and solve for [tex]S[/tex]:

[tex]S=\dfrac{117}{10}+Ce^{-t/468}[/tex]

With [tex]S(0)=0[/tex], we find [tex]C=-\dfrac{117}{10}=-11.7[/tex], so that

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

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