Question

Suppose that a large mixing tank initially holds 300 gallons of water is which 50 pounds of salt have been dissolved. Another brine solution is pumped into the tank at a rate of 3gal/min, and when the solution is well stirred, it is then pumped out at a slower rate of 2gal/min. If the concentration of the solution entering is 2lb/gal, determine a differential equation for the amount of salt A(t)in the tank at time t.

Short answer

The differential equation is$\frac{dA}{dt}=6-\frac{2A\left(t\right)}{300+t}\text{.}$

Step-by-step solution

Definition:

A differential equation is an equation with one or more derivatives of a function.

Given data:

Initially 300 gallons of water in which 50 pounds of salt into the tank at a rate of 3gal/min.

Pumped out at a slower rate is 2 gal/min.

Concentration of solution entering is 2lb/gal.

Now, use the differential equation, $\frac{dA}{dt}=R_{\text{it}}-R_{\text{oar}}\dots .\dots \left(1\right)$

Let A(t) be the amount of salt in the tank at any time t.

$$\begin{gathered} R_{\text{in}}=(\text{Rate}{)}_{\text{im}}\times (\text{concentration}{)}_{\text{in}} \\ =(3\text{gal}/\text{min})\times (2\text{lb}/\text{gal}) \\ =6\text{lb}/\text{min} \end{gathered}$$

Find rate of out flow

Since the solution is pumped out at a slower rate; it is accumulating at the rate of $(3-2)\text{gal}/\text{min}=1\text{gal}/\text{min}$.

After t minutes there are 300+t gallons of brine in the tank.

So, the volume is $V\left(t\right)=300+t$.

$$\begin{gathered} R_{\text{cut}}=(\text{Pumped rate}{)}_{\text{out}}\times (\text{concentration}{)}_{\text{out}} \\ =2\times \frac{A\left(t\right)}{V\left(t\right)} \\ =\frac{2A\left(t\right)}{300+t} \end{gathered}$$

Substitute value:

Now, substitute R_{in ,}R_0win the equation (1).

$$\begin{gathered} \frac{dA}{dt}=R_{iv}-R_{\text{ow}} \\ =6\text{lb}/\min -\frac{2A\left(t\right)}{300+t}\text{lb}/\text{min} \\ =\left(6-\frac{2A\left(t\right)}{300+t}\right)\text{lb}/\text{min} \end{gathered}$$

Hence, the required differential equation for the amount of salt in the tank at time t is

$\frac{dA}{dt}=6-\frac{2A\left(t\right)}{300+t}\text{.}$