Problem 31 - Calculus Single Variable

Question

A mixing tank contains 100 gallons of water in which 60 pounds of salt is dissolved. At time \(t=0,\) a valve is opened, and water enters the tank at the rate of 2.5 gallons per minute. An outlet pipe maintains the volume of fluid in the tank by allowing 2.5 gallons of the thoroughly mixed solution to flow out each minute. What is the weight \(m(t)\) of salt in the tank at time \(t ?\) What is \(m_{\infty}=\lim _{t \rightarrow \infty} m(t) ?\)

Step-by-Step Solution

Answer

The weight of salt at time \( t \) is \( m(t) = 60 \times e^{-\frac{1}{40}t} \), and \( m_{\infty} = 0 \).

Step 1: Set Up Salt Concentration Equation

Let \( m(t) \) be the amount of salt in the tank at time \( t \). Initially, the tank has \( m(0) = 60 \) pounds of salt. Since the concentration is uniform due to thorough mixing, the concentration of salt at any time \( t \) is \( \frac{m(t)}{100} \) pounds per gallon.

Step 2: Determine Rate of Salt Input and Output

Water is entering the tank at 2.5 gallons per minute with no additional salt, so the rate of salt input is 0. Water is leaving the tank at the same rate, 2.5 gallons per minute. The rate of salt leaving the tank is \( 2.5 \times \frac{m(t)}{100} \). This is the rate we need for the differential equation.

Step 3: Write the Differential Equation for Salt in Tank

The change in the amount of salt can be described by the differential equation \( \frac{dm}{dt} = \, \text{(rate of input)} - \, \text{(rate of output)} = 0 - 2.5 \times \frac{m(t)}{100} \). This simplifies to \( \frac{dm}{dt} = -\frac{1}{40}m(t) \).

Step 4: Solve the Differential Equation

This is a first-order linear differential equation. Its solution is in the form \( m(t) = m(0) \times e^{-\frac{1}{40}t} \), where \( m(0) = 60 \). Thus, \( m(t) = 60 \times e^{-\frac{1}{40}t} \).

Step 5: Evaluate Limit as Time Approaches Infinity

We need to find \( m_{\infty} = \lim_{t \to \infty} m(t) \). Substitute the expression found in Step 4 to get \( \lim_{t \to \infty} 60 \times e^{-\frac{1}{40}t} \). As \( t \to \infty \), \( e^{-\frac{1}{40}t} \to 0 \). Therefore, \( m_{\infty} = 60 \times 0 = 0 \) pounds of salt.

Key Concepts

Mixing Problems

Mixing problems are a classic application of differential equations, specifically focusing on how substances distribute within a solution over time. In our context, we have a tank where water flows in and out at equal rates, maintaining a constant volume of 100 gallons. However, despite the same volume, the concentration of salt changes as new water mixes with the salt solution.
These problems are not only relevant in mathematical models but also in real-life scenarios such as chemical mixing and environmental engineering.

• The incoming water has no salt, meaning it dilutes the existing solution.
• Since the mixture exits at the same rate, the system remains balanced, making it a perfect candidate to be analyzed by differential equations.

Understanding mixing problems helps us design more efficient systems involving pumps, pipes, and any circulating fluids.

Salt Concentration

Salt concentration in a solution is a measure of how much salt is dissolved in a given amount of liquid. In this problem, we start with 60 pounds of salt in 100 gallons of water, giving an initial concentration of \(\frac{60}{100} = 0.6\) pounds per gallon.
As water with no salt enters the tank and the mixture exits, this concentration changes. However, at every moment, the concentration formula \(\frac{m(t)}{100}\) helps us track how salty the water remains.
By understanding this concentration change, we can quantify how dispersing or concentrating agents like salt behave over time when subjected to inflow and outflow situations. Salt concentration dynamics are key to solving the governing differential equation, which also describes various phenomena such as drug degradation in pharmacokinetics and pollutant dispersal in the environment.

Initial Value Problem

An initial value problem in differential equations specifies the condition of the system when it starts, which helps us predict its future behavior. Here, our initial value is the amount of salt dissolved at time \(t = 0\), which is 60 pounds. This starting point, combined with the differential equation \( \frac{dm}{dt} = -\frac{1}{40}m(t) \), enables us to find an exact solution describing the salt quantity over time.The initial value \(m(0) = 60\) pounds helps us uniquely determine the function \(m(t)\). Without this initial condition, the problem would remain unsolved since differential equations often account for multiple solutions.

• Provides a concrete starting point for the differential equations.
• Ensures the uniqueness of the solution over time.

Mastering initial value problems is essential for accurately modeling and solving real-world systems predictably from their onset.

Exponential Decay

Exponential decay describes a process where a quantity decreases at a rate proportional to its current value. In this problem, the salt in the tank undergoes exponential decay, as shown by the solution \(m(t) = 60 \times e^{-\frac{1}{40}t}\).Exponential decay is common in natural processes like radioactive decay, cooling of objects, and population decline.

• Characterized by a rapid decrease initially that slows over time.
• The larger the decay constant, the faster the decrease.

As time approaches infinity, \(e^{-\frac{1}{40}t}\) trends towards zero, indicating that eventually, almost no salt remains in the tank. Understanding exponential decay is crucial for predicting how quickly quantities change in time-sensitive systems.