A large tank is filled to capacity with 400 gallons of pure water. brine containing 3 pounds of salt per gallon is pumped into the tank at a rate of 4 gal/min. the well-mixed solution is pumped out at the same rate. find the number a(t) of pounds of salt in the tank at time t. a(t) = lb

This is a problem of differential equations. Differential equations are of interest to nonmathematicians primarily because of the possibility of using them to investigate a wide variety of problems in the physical, biological, and social sciences. The main reason for this is that mathematical models and their solutions lead to equations relating the variables and parameters in the problem. So you can make predictions about how the natural process will behave in various circumstances starting from the equations.

Assuming that salt is neither created nor destroyed in the tank. Therefore, variations in the amount of salt are due solely to the flows in and out of the tank. More precisely, the rate of change of salt in the tank is:

[tex]\frac{da}{dt}=rate \ in - rate \ out[/tex]

The rate at which salt enters the tank is the concentration:

[tex]C_{i}= 3\frac{lb}{gal}[/tex]
 
times the flow rate:

[tex]Q_{i}=4 \frac{gal}{min}[/tex]

So:

[tex]rate \ in=3\frac{lb}{gal} \times 4 \frac{gal}{min}=12 \frac{lb}{min}[/tex]

To find the rate at which salt leaves the tank, we need to multiply the concentration of salt in the tank by the rate of outflow  [tex]Q_{0}\frac{gal}{min}[/tex]

Since the rates of flow in and out are equal, the volume of water in the tank remains constant at [tex]400gal[/tex] and and since the mixture is well-mixed, the concentration throughout the tank is the same, namely:

[tex]C_{2}=[\frac{a(t)}{400}]\frac{lb}{gal}[/tex]

Therefore, the rate at which salt leaves the tank is:

[tex]rate \ out=[Q_{o}\frac{a(t)}{400}]\frac{lb}{min} \\ \\ but \ Q_{0}=Q_{i}[/tex]

Therefore:

[tex]rate \ out=[Q_{i}\frac{a(t)}{400}]=4\frac{a(t)}{400}=[\frac{a(t)}{100}]\frac{lb}{min}[/tex]

Thus the differential equation governing this process is:

[tex]\frac{da}{dt}=12- \frac{a(t)}{100}[/tex]

Solving:

[tex]\frac{da}{dt}=12- \frac{a}{100} \\ \\ \therefore \frac{da}{dt}=\frac{1200-a}{100} \\ \\ \therefore da=\frac{1200-a}{100}dt \rightarrow da=\frac{-(a-1200)}{100}dt \\ \\ \therefore \frac{da}{a-1200}= -\frac{dt}{100} \\ \\ \therefore \intop_{0}^{a} \frac{da}{a-1200}=\intop_{0}^{t}-\frac{dt}{100} \\ \\ \therefore ln(a-1200)= -\frac{t}{100} \\ \\ Finally: \\ \\ a(t)=1200+ce^{-\frac{t}{100}}[/tex]

where [tex]c[/tex] is an arbitrary constant, so:

[tex]If \ a(0)=0 \ then: \\ \\ c=-1200[/tex]

Lastly, the number [tex]a(t)[/tex] of pounds of salt in the tank at time [tex]t[/tex] is:

[tex]\boxed{a(t)=1200-1200e^{-\frac{t}{100}}=1200(1-e^{-\frac{t}{100}})}[/tex]