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

The amount of salt in the tank after 40 minutes is approximately 243.75 pounds.

Step 1: Understand the System

We have a tank with 200 gallons of brine initially containing 50 pounds of salt. Brine with 2 pounds of salt per gallon is flowing in at 4 gallons per minute, and the mixture is leaving at the same rate.

Step 2: Define the Variables

Let \( S(t) \) be the amount of salt in the tank at time \( t \) (in minutes). We know \( S(0) = 50 \) because initially there are 50 pounds of salt.

Step 3: Set Up the Differential Equation

The rate of salt flowing in is \( 4 \, \text{gallons/minute} \times 2 \, \text{pounds/gallon} = 8 \, \text{pounds/minute} \). The rate of salt flowing out is \( \frac{S(t)}{200} \, \text{pounds/gallon} \times 4 \, \text{gallons/minute} = \frac{4S(t)}{200} \, \text{pounds/minute} = \frac{S(t)}{50} \, \text{pounds/minute} \).

Step 4: Write the Differential Equation

The total rate of change of salt in the tank is given by \( \frac{dS}{dt} = 8 - \frac{S(t)}{50} \).

Step 5: Solve the Differential Equation

This is a first-order linear differential equation. Rearrange it to \( \frac{dS}{dt} + \frac{S(t)}{50} = 8 \). The integrating factor is \( e^{\int \frac{1}{50} dt} = e^{t/50} \). Multiply the entire differential equation by this integrating factor, integrate both sides, and solve for \( S(t) \).

Step 6: Find the Integrating Factor and Integrate

After multiplying by the integrating factor, integrate both sides:\[ e^{t/50} S(t) = \int 8 e^{t/50} dt = 400 e^{t/50} + C \]. Solving, \( S(t) = 400 + Ce^{-t/50} \).

Step 7: Apply Initial Condition

We know \( S(0) = 50 \), so substitute to find \( C \):\[ 50 = 400 + C \Rightarrow C = -350 \].

Step 8: Find the Amount of Salt at 40 Minutes

Substitute \( t = 40 \) minutes into the equation \( S(t) = 400 - 350e^{-t/50} \):\[ S(40) = 400 - 350e^{-40/50} \].

Step 9: Calculate Final Amount of Salt

Compute \( S(40) \):\[ S(40) = 400 - 350e^{-4/5} \approx 400 - 350 \times 0.4493 \approx 243.75 \].Thus, the final amount of salt in the tank is approximately 243.75 pounds.

Key Concepts

First-Order Linear Differential Equations

First-order linear differential equations are equations that involve derivatives of a function and the function itself. These equations are called "first-order" because they contain the first derivative but not higher derivatives. They are "linear" if they can be written in the form \( a(t) \frac{dy}{dt} + b(t) y = c(t) \), where \( a(t) \), \( b(t) \), and \( c(t) \) are functions of \( t \), and \( y \) is the function we want to find.
The key to solving these equations is the integrating factor method. The integrating factor is a special function that, when multiplied by the entire differential equation, allows us to rewrite it as a perfect derivative. For the given problem, the equation \( \frac{dS}{dt} + \frac{S(t)}{50} = 8 \) represents a typical first-order linear differential equation. To solve it, we identified the integrating factor: \( e^{t/50} \), which when applied, enables integration, ultimately leading us to the solution for \( S(t) \).
The result is a function describing how the amount of salt changes over time, using the initial condition to find the particular solution.

Salt Concentration Problems

Salt concentration problems involve determining the amount or concentration of salt in a solution over time, often within a system where the solution is constantly mixed. These problems typically assume a uniform distribution of salt due to stirring, which simplifies calculations.
In our exercise, we're tracking the change in salt concentration in a tank. The inflow contains a known concentration of salt (2 pounds per gallon), and it's crucial to keep track of incoming and outgoing salt rates to set up the initial differential equation. The incoming salt rate is \( 8 \text{ pounds/minute} \), achieved by the product of the inflow rate and concentration. Meanwhile, the outgoing rate is dependent on the concentration within the tank, calculated as \( \frac{S(t)}{50} \text{ pounds/minute} \).
The overall goal is to use these rates to build an equation that models how the amount of salt changes over time, factoring the influence of both inflow and outflow. This equation gives insight into how concentration evolves, particularly under conditions of steady inflow and outflow.

Mixing Problems

Mixing problems typically involve scenarios where different substances are combined in a tank, such as brine solutions, and we're asked to track certain properties over time. In these problems, it's important to set up a model based on rates of flow into and out of the tank and how they impact the solution.
With constant stirring, it is assumed that the solution is homogenous, meaning every part of the solution has an identical concentration at any time. This homogeneity is critical as it makes the mathematical modeling simpler and allows us to apply differential equations to track changes in properties like salt concentration.
In the given exercise, with brine entering and leaving at the same rate, the process becomes cyclic and predictable, which allows the derivation of the differential equation governing the salt content. The constant inflow and outflow are balanced, and the solution to the differential equation gives the amount of salt at any point in time, eventually leading to the final calculation of 243.75 pounds of salt after 40 minutes.