How do you determine how much salt is in the tank when it is full if a 30-gallon tank initially contains 15 gallons of salt water containing 5 pounds of salt and suppose salt water containing 1 pound of salt per gallon is pumped into the top of the tank at the rate of 2 gallons per minute, while a well-mixed solution leaves the bottom of the tank at a rate of 1 gallon per minute?
1 Answer
There will be 25 pounds of salt.
Let
The rate at which salt is entering the tank is
#(2\frac{\mbox{gal}}{min})\cdot (1 \frac{\mbox{pound}}{\mbox{gal}}) = 2 \frac{\mbox{pounds}}{\mbox{min}}#
The rate at which salt is exiting the tank is
#(1\frac{\mbox{gal}}{\mbox{min}})\cdot (\frac{Q\mbox{ pounds}}{t+15\mbox{ gal}}) = \frac{Q}{t+15} \frac{\mbox{pounds}}{\mbox{min}}#
This leads to the differential equation
To solve this differential equation, write it as
#\frac{dQ}{dt}+\frac{1}{t+15}Q=2#
Then multiply both sides of this equation by the integrating factor
#\mu(x)=e^{\int \frac{1}{t+15}dt}=e^{\ln(t+15)}=t+15#
to get
#(t+15)\frac{dQ}{dt}+Q=2(t+15)=2t+30#
or (by the Product Rule)
#\frac{d}{dt}((t+15)Q)=2t+30#
Now integrate both sides to get
#(t+15)Q=t^{2}+30t+C#
so that
#Q=\frac{t^{2}+30t+C}{t+15}#
The initial condition implies that
#5=Q(0)=\frac{C}{15}#
so that
#C=75# and#Q=\frac{t^{2}+30t+75}{t+15}#
The tank will be full when
The final answer is therefore:
#Q(15)=\frac{225+450+75}{30}=\frac{750}{30}=25# pounds of salt.
Go to the following video for more help in setting it up:
Go to the following video for an integrating factor example: