Math Problem Statement

Initially 50 pounds of pollutant is added to a tank that has 300 gallons of pure water. Fluid with 2 pounds per gallon of pollutant then runs into this tank at the rate of 10 gallons per minute. The mixture is then well stirred before it flows out at the rate of 12 gallons per minute.

Solution

This problem involves modeling the pollutant concentration in the tank using a differential equation. Let's analyze the situation step-by-step:

  1. Initial Setup:

    • Tank initially has 300 gallons of water and 50 pounds of pollutant.
    • Pollutant inflow: 2 pounds per gallon of pollutant solution, flowing in at 10 gallons per minute.
    • The tank is well stirred, and the mixture flows out at a rate of 12 gallons per minute.
  2. Key Relationships:

    • Rate of pollutant inflow: Rate in=(concentration of pollutant in inflow)×(rate of inflow)=2×10=20pounds per minute.\text{Rate in} = \text{(concentration of pollutant in inflow)} \times \text{(rate of inflow)} = 2 \times 10 = 20 \, \text{pounds per minute}.
    • Rate of pollutant outflow: The outflow rate depends on the concentration of pollutant in the tank at any given time. If A(t)A(t) is the amount of pollutant (in pounds) in the tank at time tt, and the tank volume is V(t)V(t), then: Rate out=A(t)V(t)×(rate of outflow)=A(t)V(t)×12.\text{Rate out} = \frac{A(t)}{V(t)} \times \text{(rate of outflow)} = \frac{A(t)}{V(t)} \times 12.
    • Volume change in the tank: The inflow rate is 10 gallons/min, and the outflow rate is 12 gallons/min. Thus, the tank volume decreases over time: V(t)=3002t(gallons).V(t) = 300 - 2t \, \text{(gallons)}.
  3. Differential Equation: Using the principle of rate of change of pollutant in the tank: dAdt=Rate inRate out.\frac{dA}{dt} = \text{Rate in} - \text{Rate out}. Substituting the expressions for rate in and rate out: dAdt=20A(t)3002t×12.\frac{dA}{dt} = 20 - \frac{A(t)}{300 - 2t} \times 12. This equation models the change in the pollutant amount A(t)A(t) over time.

  4. Boundary Condition: At t=0t = 0, the initial amount of pollutant is A(0)=50A(0) = 50.

Would you like to proceed with solving this differential equation explicitly or discuss it further?

Related Questions:

  1. How does the tank's decreasing volume affect the pollutant concentration over time?
  2. Can we solve the differential equation numerically if an analytical solution is complex?
  3. What is the critical time when the tank would be empty (V(t)=0V(t) = 0)?
  4. How would changing the inflow or outflow rates affect the dynamics of the pollutant concentration?
  5. What happens if the pollutant inflow concentration changes over time?

Tip:

Always check if the tank will empty before solving such problems—it significantly changes the scenario when ( V(t

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Math Problem Analysis

Mathematical Concepts

Differential equations
Rate of change
Mixing problems

Formulas

Rate of pollutant inflow: (Concentration of pollutant in inflow) × (Rate of inflow)
Rate of pollutant outflow: (Concentration of pollutant in tank) × (Rate of outflow)
Volume of tank: Initial volume - (Outflow rate - Inflow rate) × Time

Theorems

Law of Conservation of Mass

Suitable Grade Level

College Level or Advanced High School (Calculus-Based Problems)