Question

A tank initially contains 90 gallons of water and 60 ounces of salt. Water containing a salt concentration of 0.2 ounces per gallon flows into the tank at a rate of 9 gallons per minute, and the well-stirred mixture leaves the tank at a rate of 5 gallons per minute. Let Q(t) be the amount of salt (in ounces) at t minutes after this process started.

Answer 1

The rate of salt entering the tank is given by the concentration of salt in the incoming water multiplied by the rate at which water is entering the tank. The rate of salt leaving the tank is given by the concentration of salt in the tank at that time multiplied by the rate at which water is leaving the tank. The differential equation representing the rate of change of salt in the tank is dQ/dt = 1.8 - (Q(t) * 5).

Step-by-step explanation:

The amount of salt in the tank at any given time can be represented by a differential equation. Let's consider the rate at which salt is entering and leaving the tank. The rate of salt entering the tank is given by the concentration of salt in the incoming water (0.2 ounces per gallon) multiplied by the rate at which water is entering the tank (9 gallons per minute). This gives us a rate of (0.2 * 9) = 1.8 ounces per minute. The rate of salt leaving the tank is given by the concentration of salt in the tank at that time (Q(t) ounces) multiplied by the rate at which water is leaving the tank (5 gallons per minute). This gives us a rate of (Q(t) * 5) ounces per minute.

So, the differential equation representing the rate of change of salt in the tank is dQ/dt = 1.8 - (Q(t) * 5).

We can solve this differential equation using separation of variables to find an expression for Q(t).