How to model the amount of salt in the tank at any time, and after a specific amount of time has passed
Example
A tank contains ???1,500??? L of water and ???20??? kg of dissolved salt. Fresh water is entering the tank at ???15??? L/min (the solution stays perfectly mixed), and the solution drains at a rate of ???10??? L/min. How much salt is in the tank at ???t??? minutes and at ???10??? minutes?
We’ll start with the mixing problem formula
???\frac{dy}{dt}=C_1r_1-C_2r_2???
In this problem, we’re interested in the concentration of salt in the tank.
???C_1=0??? kg/L because no salt is being added into the tank.
???r_1=15??? L/min because this is the rate at which water is entering the tank
???C_2=y/(1,500+5t)??? kg/L because we’re not sure how much salt is leaving the tank, but we know the initial amount of water is ???1,500??? L, and we’re adding ???15-10=5??? L every minute
???r_2=10??? L/min because this is the rate at which the solution is leaving the tank
If we plug all these values into the formula, we get
???\frac{dy}{dt}=(0)(15)-\left(\frac{y}{1,500+5t}\right)(10)???
???\frac{dy}{dt}=-\frac{10y}{1,500+5t}???
???\frac{dy}{dt}=-\frac{2y}{300+t}???
Now we’ll separate the variables.
???dy=-\frac{2y}{300+t}\ dt???
???\frac{1}{y}\ dy=-\frac{2}{300+t}\ dt???
With the variables separated, we’ll integrate both sides of the equation.
???\int\frac{1}{y}\ dy=\int-\frac{2}{300+t}\ dt???
???\ln{|y|}+C_1=-2\ln{|300+t|}+C_2???
???\ln{|y|}=-2\ln{|300+t|}+C_2-C_1???
Raise both sides to the base ???e??? in order to eliminate the natural log.
???e^{\ln{|y|}}=e^{-2\ln{|300+t|}+C_2-C_1}???
???|y|=e^{C_2-C_1}e^{-2\ln{|300+t|}}???
???y=\pm e^{C_2-C_1}e^{\ln{(|300+t|)^{-2}}}???
???y=\pm e^{C_2-C_1}(|300+t|)^{-2}???
The ???\pm e^{C_2-C_1}??? is still just a constant, so we can simplify.
???y=C(|300+t|)^{-2}???
Time ???t??? will always be positive (it’s nonsensical to have “negative time”), so ???300+t??? will always be positive, and we can remove the absolute value bars from the ???|300+t|???.
???y=C(300+t)^{-2}???
???y=\frac{C}{(300+t)^2}???
We were told that initially ???20??? kg of dissolved salt existed in the tank, so ???y(0)=20???.
???20=\frac{C}{(300+0)^2}???
???20=\frac{C}{90,000}???
???C=1,800,000???
Plugging this back into the general solution, we get
???y=\frac{1,800,000}{(300+t)^2}???
This is the equation that models the amount of salt in the tank at ???t??? minutes. If we want to figure out how much salt is in the tank after ???5??? minutes, we just plug ???5??? in for ???t???. If we want to figure out how much salt is in the tank after ???20??? minutes, we just plug ???20??? in for ???t???.
We’ve also been asked in this problem to find the amount of salt in the tank after ???10??? minutes. Plugging ???10??? in for ???t???, we get
???y=\frac{1,800,000}{(300+10)^2}???
???y=\frac{1,800,000}{310^2}???
???y=\frac{1,800,000}{96,100}???
???y=\frac{18,000}{961}???
???y\approx18.73???
After ???10??? minutes, there’s ???18.73??? kg of salt in the tank.