Respuesta :
Answer:
[tex]A(t)=300-290e^{-\frac{t}{60}}[/tex]
Step-by-step explanation:
The volume of fluid in the tank = 300 Liters
Initial amount of Salt in the tank, A(0)=10 grams
Change in 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]R_{in}=(1\frac{gram}{L})( 5\frac{L}{min})=5\frac{grams}{min}[/tex]
Rate Out =(concentration of salt in outflow)(output rate of brine)
[tex]R_{out}=(\frac{A(t)}{300})( 5\frac{L}{min})=\frac{A(t)}{60}[/tex]
Therefore:
[tex]\dfrac{dA}{dt}=5-\dfrac{A(t)}{60}\\$Rearranging, we have:\\\dfrac{dA}{dt}+\dfrac{A(t)}{60}=5[/tex]
We solve the resulting linear differential equation for A(t)
[tex]\text{The integrating factor: } e^{\int \frac{1}{60}dt} =e^{\frac{t}{60}}\\$Multiplying by the integrating factor all through\\\dfrac{dA}{dt}e^{\frac{t}{60}}+\dfrac{A}{60}e^{\frac{t}{60}}=5e^{\frac{t}{60}}\\(Ae^{\frac{t}{60}})'=5e^{\frac{t}{60}}[/tex]
Taking the integral of both sides
[tex]\int(Ae^{\frac{t}{60}})'=\int 5e^{\frac{t}{60}} dt\\Ae^{\frac{t}{60}}=5*60e^{\frac{t}{60}}+C, $(C a constant of integration)\\Ae^{\frac{t}{60}}=300e^{\frac{t}{60}}+C\\$Divide all through by e^{\frac{t}{60}}\\A(t)=300+Ce^{-\frac{t}{60}}[/tex]
Recall that when t=0, A(t)=10 grams (our initial condition)
[tex]10=300+Ce^{-\frac{0}{60}}\\10-300=Ce^0\\C=-290[/tex]
Therefore, the number A(t) of grams of salt in the tank at time t is:
[tex]A(t)=300-290e^{-\frac{t}{60}}[/tex]
