Answer:
[tex]\therefore \quad \frac{d A}{d t}=6-\frac{2 A(t)}{300+t}[/tex]
Explanation:
[tex]\because \quad \frac{d A}{d t}=R_{i n}-R_{\text {out }}[/tex]
Thus, we find [tex]R_{i n}$ and $R_{\text {out first }}[/tex]
[tex]\because R_{i n}=( concentration of salt in inflow ) \cdot[/tex] (input rate of brine)
[tex]\therefore \quad R_{i n}=(2 \mathrm{lb} / \mathrm{gal}) \cdot(3 \mathrm{gal} / \mathrm{min})=6 \mathrm{lb} / \mathrm{min}[/tex]
Since the solution is pumped out at a slower rate, thus it is accumulating
at the rate of [tex](3-2)=1 \mathrm{gal} / \mathrm{min}[/tex]
Thus, after t minutes there will be [tex]300+t[/tex] gallons in tank
[tex]\because R_{\text {out }}=[/tex] (concentration of salt in outflow) \cdot (output rate of brine)
[tex]\therefore \quad R_{o u t}=\left(\frac{A(t)}{300+t} \mathrm{lb} / \mathrm{gal}\right) \cdot(2 \mathrm{gal} / \mathrm{min})=\frac{2 A(t)}{300+t} \mathrm{lb} / \mathrm{min}[/tex]
Now, we substitute by these results in the [tex]\mathrm{DE}[/tex] to get
[tex]\therefore \quad \frac{d A}{d t}=6-\frac{2 A(t)}{300+t}[/tex]
