Respuesta :
Answer:
[tex] A(40)= \frac{-200}{10+40} +4 (10 +40)=-4+200 = 196 [/tex]
Step-by-step explanation:
For this case the solution flows at a rate of 2L/min and leaves at 1L/min. So then we can conclude the volume is given by [tex] V= 10 +t[/tex]
Since the initial volume is 10 L and the volume increase at a rate of 1L/min.
For this case we can define A as the concentration for the salt in the container. And for this case we can set up the following differential equation:
[tex] \frac{dA}{dt}= 4 \frac{gr}{L} *2 \frac{L}{min} - \frac{A}{10+t}[/tex]
Because at the begin we have a concentration of 8 gr/L and would be decreasing at a rate of [tex] \frac{A}{10+t}[/tex]
So then we can reorder the differential equation like this:
[tex] \frac{dA}{dt} +\frac{A}{10+t} =8[/tex]
We find the solution using the integration factor:
[tex] \mu = -\int \frac{1}{10+t} dt = -ln(10+t)[/tex]
And then the solution would be given by:
[tex] A = e^{-ln (10+t)} (\int e^{\int \frac{1}{10+t} dt})[/tex]
And if we simplify this we got:
[tex] A= \frac{1}{10+t} (c + \int (10 +t) 8 dt)[/tex]
And after do the integral we got:
[tex] A= \frac{c}{10+t} +4 (10 +t) [/tex]
And using the initial condition t=0 A= 20 we have this:
[tex] 20 = \frac{c}{10} +40[/tex]
[tex] c= -200[/tex]
So then we have this function for the solution of A:
[tex] A= \frac{-200}{10+t} +4 (10 +t) [/tex]
And now replacinf t= 40 we got:
[tex] A(40)= \frac{-200}{10+40} +4 (10 +40)=-4+200 = 196 [/tex]
