Problem 16 - Calculus Early Transcendentals: Pearson New International Edition

Question

A tank initially contains 200 gallons of brine, with 50 pounds of salt in solution. Brine containing 2 pounds of salt per gallon is entering the tank at the rate of 4 gallons per minute and is flowing out at the same rate. If the mixture in the tank is kept uniform by constant stirring, find the amount of salt in the tank at the end of 40 minutes.

Step-by-Step Solution

Answer

After 40 minutes, there are approximately 242.75 pounds of salt in the tank.

Step 1: Understand the problem and its components

We are asked to find the amount of salt in a tank at the end of 40 minutes. The tank initially has 200 gallons of brine with 50 pounds of salt. Brine with 2 pounds of salt per gallon enters the tank at 4 gallons per minute, and the same amount exits the tank.

Step 2: Define the variables and initial conditions

Let \( S(t) \) be the amount of salt (in pounds) at time \( t \), with \( S(0) = 50 \) pounds. The rate of salt entering is \( 2 \cdot 4 = 8 \) pounds per minute. Since the outgoing brine has the same concentration as inside, the outgoing rate is \( \frac{S(t)}{200} \times 4 \) pounds per minute.

Step 3: Set up a differential equation

The change in amount of salt is given by the difference in the rate of salt entering and leaving: \[ \frac{dS}{dt} = 8 - \frac{4S}{200} = 8 - \frac{S}{50}. \]

Step 4: Solve the differential equation

Rearrange the differential equation: \[ \frac{dS}{dt} + \frac{S}{50} = 8. \] This is a linear first-order differential equation. Use an integrating factor to solve it. The integrating factor is \( e^{t/50} \). Multiply through by the integrating factor:\[ e^{t/50} \frac{dS}{dt} + \frac{S}{50} e^{t/50} = 8 e^{t/50}. \]

Step 5: Integrate both sides

The left-hand side is the derivative of \( S(t) e^{t/50} \):\[ \frac{d}{dt}(S e^{t/50}) = 8 e^{t/50}. \]Integrate both sides with respect to \( t \):\[ S e^{t/50} = 8 \int e^{t/50} dt = 400 e^{t/50} + C. \]

Step 6: Solve for \( S(t) \) and apply initial conditions

Solve for \( S(t) \):\[ S(t) = 400 + C e^{-t/50}. \] Use \( S(0) = 50 \) to find \( C \):\[ 50 = 400 + C, \]\[ C = -350. \]Thus, \( S(t) = 400 - 350 e^{-t/50}. \)

Step 7: Calculate \( S(40) \)

Substitute \( t = 40 \):\[ S(40) = 400 - 350 e^{-40/50} = 400 - 350 e^{-0.8}. \]Evaluate \( e^{-0.8} \) and compute the final amount of salt,using \( e^{-0.8} \approx 0.4493 \),\[ S(40) \approx 400 - 350 \times 0.4493 \approx 400 - 157.255 \approx 242.745. \]

Key Concepts

Brine Tank Problem

The brine tank problem is a classic example of how differential equations are used to model real-world systems. In these problems, a tank initially contains a certain volume of fluid, known as brine, which is a solution of salt and water. As time progresses, the tank is subjected to an influx and outflux of brine, usually at equal rates to maintain balance within the tank.

In the exercise, we start with a tank containing 200 gallons of brine, with an initial salt content of 50 pounds. Brine enriched with 2 pounds of salt per gallon then flows into the tank at a rate of 4 gallons per minute. The same amount exits the tank per minute. Thus, the system's conditions highlight the need to monitor and calculate the dynamic net change in salt content over time. Because the tank is uniformly stirred, the concentration of salt inside remains constant throughout this process, making this a perfect scenario for modeling with differential equations.

Integrating Factor

The method of integrating factors is a powerful tool for solving linear first-order differential equations. In this context, we aim to solve equations of the form \( \frac{dS}{dt} + P(t) S = Q(t) \), where \( P(t) \) and \( Q(t) \) are functions of time.

To solve such equations, an integrating factor \( \mu(t) \) is applied, which when multiplied throughout the equation, transforms it into one that is easily integrable. We calculate \( \mu(t) = e^{\int P(t) \, dt} \). In our brine tank problem, the differential equation \( \frac{dS}{dt} + \frac{S}{50} = 8 \) is solved using this approach.

The integrating factor in this case becomes \( e^{t/50} \). Upon multiplying this factor across the equation, the left side simplifies to the derivative of a product, making it straightforward to integrate both sides. This results in a solution expressing \( S(t) \) as the amount of salt at any given time \( t \). This systematic approach highlights how differential equations model changes in systems efficiently.

First-order Differential Equation

A first-order differential equation is characterized by the presence of the first derivative of a function, without any higher-order derivatives. These types of equations often model systems where a quantity changes at a rate proportional to its current value and other influencing factors.

In this particular brine tank problem, the rate of change of the salt content \( S(t) \) within the tank is described as \( \frac{dS}{dt} = 8 - \frac{S}{50} \). The first term, 8, represents the constant rate at which salt enters the tank, while the second term, \( \frac{S}{50} \), signifies the concentration-dependent rate of salt leaving the tank.

Transforming this into a more solvable format involves rearranging the terms, ultimately gearing up for the application of an integrating factor. This allows us to obtain \( S(t) \) by solving for the time-dependent solution, illustrating the powerful capability of first-order differential equations in predicting system behavior over time.

Rate of Change in Solutions

The rate of change in a solution refers to how quickly the concentration of a particular component within a mixture changes over time. In the context of the brine tank problem, this rate of change is crucial in determining the dynamic behavior of the salt concentration in the brine.

To be precise, the problem calculates how the salt content evolves as brine flows in and out of a tank, under the influence of constant stirring and equal rates of inflow and outflow. Here, the rate is governed by a differential equation \( \frac{dS}{dt} = 8 - \frac{S}{50} \). This equation succinctly captures two dynamics:

• The entry of salt into the tank at a constant rate of 8 pounds per minute.
• The removal of salt, a process dependent on the current concentration \( S(t)/200 \), and hence influences the outgoing rate.

The balance of these rates determines how the salt concentration adjusts over time. Integrating this differential equation reveals how the salt content will behave after 40 minutes, making it a beautifully illustrative example of how differential equations direct predictions on the evolution of solutions in chemical and physical systems.