Answer:
a) ![y(t)=50000-49990e^{\frac{-2t}{25}}](https://tex.z-dn.net/?f=y%28t%29%3D50000-49990e%5E%7B%5Cfrac%7B-2t%7D%7B25%7D%7D)
b) ![31690.7 g/L](https://tex.z-dn.net/?f=31690.7%20g%2FL)
Step-by-step explanation:
By definition, we have that the change rate of salt in the tank is
, where
is the rate of salt entering and
is the rate of salt going outside.
Then we have,
, and
![R_{o}=40\frac{L}{min}*\frac{y}{500} \frac{g}{L}=\frac{2y}{25}\frac{g}{min}](https://tex.z-dn.net/?f=R_%7Bo%7D%3D40%5Cfrac%7BL%7D%7Bmin%7D%2A%5Cfrac%7By%7D%7B500%7D%20%5Cfrac%7Bg%7D%7BL%7D%3D%5Cfrac%7B2y%7D%7B25%7D%5Cfrac%7Bg%7D%7Bmin%7D)
So we obtain.
, then
, and using the integrating factor
, therefore
, we get
, after integrating both sides
, therefore
, to find
we know that the tank initially contains a salt concentration of 10 g/L, that means the initial conditions
, so ![10= 50000+Ce^{\frac{-0*2}{25}}](https://tex.z-dn.net/?f=10%3D%2050000%2BCe%5E%7B%5Cfrac%7B-0%2A2%7D%7B25%7D%7D)
![10=50000+C\\C=10-50000=-49990](https://tex.z-dn.net/?f=10%3D50000%2BC%5C%5CC%3D10-50000%3D-49990)
Finally we can write an expression for the amount of salt in the tank at any time t, it is ![y(t)=50000-49990e^{\frac{-2t}{25}}](https://tex.z-dn.net/?f=y%28t%29%3D50000-49990e%5E%7B%5Cfrac%7B-2t%7D%7B25%7D%7D)
b) The tank will overflow due Rin>Rout, at a rate of
, due we have 500 L to overflow
, so we can evualuate the expression of a)
, is the salt concentration when the tank overflows