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A tank contains 30 lb of salt dissolved in 300 gallons of water. a brine solution is pumped into the tank at a rate of 3 gal/min; it mixes with the solution there, and then the mixture is pumped out at a rate of 3 gal/min. determine a(t), the amount of salt in the tank at time t, if the concentration of salt in the inflow is variable and given by cin(t) = 2 + sin(t/4) lb/gal.

Respuesta :

LammettHash
[tex]A'(t)=(\text{flow rate in})(\text{inflow concentration})-(\text{flow rate out})(\text{outflow concentration})[/tex]
[tex]\implies A'(t)=\dfrac{3\text{ gal}}{1\text{ min}}\cdot\left(2+\sin\dfrac t4\right)\dfrac{\text{lb}}{\text{gal}}-\dfrac{3\text{ gal}}{1\text{ min}}\cdot\dfrac{A(t)\text{ lb}}{300+(3-3)t\text{ gal}}[/tex]
[tex]A'(t)+\dfrac1{100}A(t)=6+3\sin\dfrac t4[/tex]

We're given that [tex]A(0)=30[/tex]. Multiply both sides by the integrating factor [tex]e^{t/100}[/tex], then

[tex]e^{t/100}A'(t)+\dfrac1{100}e^{t/100}A(t)=6e^{t/100}+3e^{t/100}\sin\dfrac t4[/tex]
[tex]\left(e^{t/100}A(t)\right)'=6e^{t/100}+3e^{t/100}\sin\dfrac t4[/tex]
[tex]e^{t/100}A(t)=600e^{t/100}-\dfrac{150}{313}e^{t/100}\left(25\cos\dfrac t4-\sin\dfrac t4\right)+C[/tex]
[tex]A(t)=600-\dfrac{150}{313}\left(25\cos\dfrac t4-\sin\dfrac t4\right)+Ce^{-t/100}[/tex]

Given that [tex]A(0)=30[/tex], we have

[tex]30=600-\dfrac{150}{313}\cdot25+C\implies C=-\dfrac{174660}{313}\approx-558.02[/tex]

so the amount of salt in the tank at time [tex]t[/tex] is

[tex]A(t)\approx600-\dfrac{150}{313}\left(25\cos\dfrac t4-\sin\dfrac t4\right)-558.02e^{-t/100}[/tex]

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