Question

a tank initially contains 200 gal of brine in which 50 lb of salt are dissolved. a brine containing 5 lb/gal of salt runs into the tank at the rate of 3 gal/min. the mixture is kept uniform by stirring and flows out of the tank at the rate of 2 gal/min. let y represent the amount of salt at time t. complete parts a through e

a. at what rate (pounds per minute) does salt enter the tank at time t?
15 lb/min

b. what is the volume of brine in the tank at time t?
(200 + t) gal

c. at what rate (pounds per minute) does salt leave the tank at time t?
\\(\frac{2y}{200 + t}\\) lb/min

d. write down and solve the initial value problem describing the mixing process.
\\(\frac{dy}{dt} = \square\\), y(0) = 50

Explanation:

Step1: Determine the rate of salt entering

The rate of salt entering the tank is calculated by multiplying the concentration of the incoming brine by the rate of inflow.
The concentration of incoming brine is \(5\) lb/gal and the inflow rate is \(3\) gal/min. So, the rate of salt entering \(=5\times3 = 15\) lb/min.

Step2: Determine the rate of salt leaving

The volume of brine in the tank at time \(t\) is \(V(t)=200 + t\) (from part b). The concentration of salt in the tank at time \(t\) is \(\frac{y}{200 + t}\) lb/gal. The outflow rate is \(2\) gal/min. So, the rate of salt leaving \(=2\times\frac{y}{200 + t}=\frac{2y}{200 + t}\) lb/min.

Step3: Set up the differential equation

Using the principle \(\frac{dy}{dt}=\text{Rate in}-\text{Rate out}\), we substitute the values from Step1 and Step2.
\(\frac{dy}{dt}=15-\frac{2y}{200 + t}\)

Answer:

\(\frac{dy}{dt}=15-\frac{2y}{200 + t}\)