Answer:
The answer is below
Step-by-step explanation:
a) The maximum capacity of he tank is 6 L and initially it contains 11 mg of salt dissolved in 3 L of water. Solution enters the tank at a rate of 3 L/hr, therefore in x hours, the amount of water that have entered the tank = 3x.
Solution also leaves the tank at a rate of 2L/hr, therefore in x hours, the amount of water that have left the tank = 2x
Hence the amount of water present in the tank at x hours is given as:
3 + 3x - 2x = 3 + x
The time taken to full the tank can be gotten from:
3 + x = 6
x = 6 - 3
x = 3 hr
b)
[tex]\frac{dQ}{dx}=3-\frac{2Q}{3+x}\\ \\\frac{dQ}{dx}+\frac{2Q}{3+x}=3\\\\let\ u'=\frac{2u}{3+x}\\\\\frac{u'}{u}=\frac{2Q}{3+x}\\\\ln(u)=2ln(3+x)\\\\u=(3+x)^2\\\\(3+x)^2Q]'=3(3+x)^2\\\\(3+x)^2Q=(3+x)^3+c\\\\Q(0)=11\\\\(3+0)^2(11)=(3+0)^3+c\\\\x=72\\\\Q=x+3+\frac{72}{(x+3)^2}\\ \\Q(3)=3+3+\frac{72}{(3+3)^2}=8\ mg[/tex]
8 mg/ 6 L = 4/3 mg/L
