Problem 9 - Advanced Engineering Mathematics

Question

Suppose that a large mixing tank initially holds 300 gallons of water in which 50 pounds of salt has been dissolved. Pure water is pumped into the tank at a rate of \(3 \mathrm{gal} / \mathrm{min}\), and when the solution is well stirred, it is pumped out at the same rate. Determine a differential equation for the amount \(A(t)\) of salt in the tank at time \(t\). What is \(A(0)\) ?

Step-by-Step Solution

Answer

The differential equation is \(\frac{dA}{dt} = -\frac{1}{100}A(t)\) with initial condition \(A(0) = 50\).

Step 1: Understanding the Initial Condition

The tank initially has 300 gallons of water containing 50 pounds of salt. So, the initial condition can be stated as \( A(0) = 50 \) pounds.

Step 2: Analyzing the Inflow and Outflow

Pure water flows into the tank at \(3\ \mathrm{gal/min}\) without changing the salt content since it contains no salt. Meanwhile, a thoroughly mixed solution goes out of the tank at the same rate \(3\ \mathrm{gal/min}\). Therefore, the volume of the solution remains 300 gallons at all times.

Step 3: Setting up the Differential Equation

\(A(t)\) represents the amount of salt in the tank at time \(t\). The concentration of salt at any time \(t\) can be given by \(\frac{A(t)}{300}\ \mathrm{pounds/gal}\). The outbound rate of salt is \(3\ \mathrm{gal/min}\) multiplied by \(\frac{A(t)}{300}\ \mathrm{pounds/gal}\). Thus, \(\frac{dA}{dt} = -3\left(\frac{A(t)}{300}\right)\).

Step 4: Simplifying the Differential Equation

Simplify the equation \(\frac{dA}{dt} = -\frac{3}{300}A(t)\) to \(\frac{dA}{dt} = -\frac{1}{100}A(t)\). This is the differential equation describing the rate of change of salt in the tank with respect to time.

Step 5: Stating Solution Initial Condition

The initial condition, which helps in solving this differential equation, is \(A(0) = 50\), meaning there are 50 pounds of salt initially in the tank.

Key Concepts

Initial Condition

In problems involving differential equations, the initial condition is crucial. It provides the starting point for the system being analyzed. In this exercise, the tank begins with 300 gallons of water and 50 pounds of salt. This gives us the initial condition, expressed mathematically as \(A(0) = 50\) pounds of salt.

• Initial condition helps define the particular solution to a differential equation.
• Without an initial condition, many solutions may exist, creating ambiguity.

In real-world problems like our mixing tank scenario, knowing the initial amount of a substance ensures that predictions or calculations stay accurate over time.

Rate of Change

The rate of change in a differential equation describes how a particular quantity varies with time. In mixing problems, it often indicates how the concentration of a substance, such as salt, changes owing to inflow and outflow rates.In our problem, since pure water enters and the solution leaves at the same rate, the system stays consistent in volume. The rate of change of salt is described by the derivative \(\frac{dA}{dt}\). This represents the speed at which salt is leaving the tank.

• The rate of change can be positive or negative. Here, it's negative because the salt is decreasing.
• Understanding rates of change helps predict future states of the system under current conditions.

The differential equation for this problem simplifies to \(\frac{dA}{dt} = -\frac{1}{100}A(t)\), capturing how salt concentration decreases over time.

Concentration

Concentration is a measure of how much of a solute is present in a solution. In this scenario, it tells us how much salt per gallon of water there is at any given time \(t\). Initially, the concentration of salt is given by \(\frac{A(0)}{300}\). As pure water continues to flow in and the mixed solution is pumped out, this concentration changes.

• Concentration varies due to mixing and the rates of inflow and outflow.
• It's crucial for determining how quickly a solute can be removed or reduced in systems like the mixing tank.

For the given differential equation, the concentration impacts the rate of salt leaving the tank, making it pivotal in solving such problems.

Mixing Problems

Mixing problems in differential equations involve understanding how different substances mix over time. They often include inflow and outflow processes that affect the concentration of a solute within a solution.In this specific exercise:

• Pure water adds to the tank, while mixed water with salt exits at the same rate.
• We assume thorough mixing so the solution composition is uniform throughout.

This scenario leads us to analyze how the salt concentration reduces over time. The differential equation \(\frac{dA}{dt} = -\frac{1}{100}A(t)\), illustrates how salt is removed consistently. Understanding mixing problems aids in various applications like chemical processes, environmental modeling, and industrial mixing systems.