Mathematics, 13.07.2020 19:01 Lia1love
A tank contains 40 lb of salt dissolved in 500 gallons of water. A brine solution is pumped into the tank at a rate of 5 gal/min; it mixes with the solution there, and then the mixture is pumped out at a rate of 5 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)
Answers: 1
![A(t)=500C_{in}(t)+[40-500C_{in}(t)]\cdot e^{-\frac{t}{100}}](/tpl/images/0703/8565/dc2ce.png)
![\int(Ae^{\frac{t}{100}})'=\int [5C_{in}(t)e^{\frac{t}{100}}]dt\\Ae^{\frac{t}{100}}=5*100C_{in}(t)e^{\frac{t}{100}}+C, $(C a constant of integration)\\Ae^{\frac{t}{100}}=500C_{in}(t)e^{\frac{t}{100}}+C\\$Divide all through by e^{\frac{t}{100}}\\A(t)=500C_{in}(t)+Ce^{-\frac{t}{100}}](/tpl/images/0703/8565/cfa3f.png)
![A(t)=500C_{in}(t)+Ce^{-\frac{t}{100}}\\40=500C_{in}(t)+Ce^{-\frac{0}{100}}\\C=40-500C_{in}(t)\\$Therefore, the amount of salt in the tank at any time t is:\\\\A(t)=500C_{in}(t)+[40-500C_{in}(t)]\cdot e^{-\frac{t}{100}}](/tpl/images/0703/8565/4ce82.png)
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