Respuesta :
Answer:
[tex]A(t)=500C_{in}(t)+[40-500C_{in}(t)]\cdot e^{-\frac{t}{100}}[/tex]
Step-by-step explanation:
Volume of water in the Tank =500 gallons
Let A(t) be the amount of salt in the tank at time t.
Initially, the tank contains 40 lbs of salt, therefore:
A(0)=40 lbs
Rate of change of the amount of Salt in the Tank
[tex]\dfrac{dA}{dt}=R_{in}-R_{out}[/tex]
Rate In=(concentration of salt in inflow)(input rate of brine)
[tex]=(C_{in}(t))( 5\frac{gal}{min})\\=5C_{in}(t)\frac{lbs}{min}[/tex]
Rate Out=(concentration of salt in outflow)(output rate of brine)
[tex]=(\frac{A(t)}{500})( 5\frac{gal}{min})=\frac{A}{100}[/tex]
Therefore:
[tex]\dfrac{dA}{dt}=5C_{in}(t)-\dfrac{A}{100}[/tex]
We then solve the resulting differential equation by separation of variables.
[tex]\dfrac{dA}{dt}+\dfrac{A}{100}=5C_{in}(t)\\$The integrating factor: e^{\int \frac{1}{100}}dt =e^{\frac{t}{100}}\\$Multiplying by the integrating factor all through\\\dfrac{dA}{dt}e^{\frac{t}{100}}+\dfrac{A}{100}e^{\frac{t}{100}}=5C_{in}(t)e^{\frac{t}{100}}\\(Ae^{\frac{t}{100}})'=5C_{in}(t)e^{\frac{t}{100}}[/tex]
Taking the integral of both sides
[tex]\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}}[/tex]
Recall that when t=0, A(t)=40 lbs (our initial condition)
[tex]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}}[/tex]
