Problem 15 - Calculus Early Transcendentals: Pearson New International Edition

Question

A tank contains 20 gallons of a solution, with 10 pounds of chemical \(A\) in the solution. At a certain instant, we begin pouring in a solution containing the same chemical in a concentration of 2 pounds per gallon. We pour at a rate of 3 gallons per minute while simultaneously draining off the resulting (well- stirred) solution at the same rate. Find the amount of chemical \(\mathrm{A}\) in the tank after 20 minutes.

Step-by-Step Solution

Answer

Approximately 38.506 pounds of chemical A are in the tank after 20 minutes.

Step 1: Determine the Initial Setup

The tank contains 20 gallons of solution initially, with 10 pounds of chemical A. The concentration of the incoming solution is 2 pounds per gallon, and it is poured in at 3 gallons per minute. The outgoing solution is also drained at 3 gallons per minute, meaning the total volume in the tank remains constant at 20 gallons.

Step 2: Set Up the Differential Equation

Let \( y(t) \) represent the amount of chemical A in the tank at time \( t \) minutes. The rate at which chemical A flows in is \( 3 \times 2 = 6 \) pounds per minute. The concentration of chemical A in the tank at time \( t \) is \( \frac{y(t)}{20} \), and the outflow rate is \( 3 \) gallons per minute, leading to a loss rate of \( 3 \times \frac{y(t)}{20} = \frac{3y(t)}{20} \) pounds per minute. Thus, the differential equation is: \( \frac{dy}{dt} = 6 - \frac{3}{20}y \).

Step 3: Solve the Differential Equation

The differential equation is separable: \( \frac{dy}{dt} + \frac{3}{20}y = 6 \). It can also be treated as a linear first-order equation. Let \( P(t) = \frac{3}{20} \) and \( Q(t) = 6 \). The integrating factor is \( \mu(t) = e^{\int \frac{3}{20} dt} = e^{\frac{3}{20}t} \). Multiplying through by \( \mu(t) \), we get \( e^{\frac{3}{20}t} \frac{dy}{dt} + e^{\frac{3}{20}t} \frac{3}{20}y = 6e^{\frac{3}{20}t} \). This simplifies to \( \frac{d}{dt}[ye^{\frac{3}{20}t}] = 6e^{\frac{3}{20}t} \). Integrate both sides to obtain: \( ye^{\frac{3}{20}t} = 40e^{\frac{3}{20}t} + C \). Divide by \( e^{\frac{3}{20}t} \) to solve for \( y(t) \): \( y(t) = 40 + Ce^{-\frac{3}{20}t} \).

Step 4: Apply Initial Condition

Initially, when \( t = 0 \), there are 10 pounds of chemical A in the tank, thus \( y(0) = 10 \). Substitute \( y(0) = 10 \) into the equation: \( 10 = 40 + C \rightarrow C = -30 \). So the solution becomes \( y(t) = 40 - 30e^{-\frac{3}{20}t} \).

Step 5: Compute the Amount After 20 Minutes

Substitute \( t = 20 \) into \( y(t) = 40 - 30e^{-\frac{3}{20}t} \) to find the amount after 20 minutes: \( y(20) = 40 - 30e^{-3/20 \times 20} = 40 - 30e^{-3} \). Compute the numerical value of \( e^{-3} \) to get the amount. If \( e^{-3} \approx 0.0498 \), then \( y(20) \approx 40 - 30 \times 0.0498 \approx 40 - 1.494 \approx 38.506 \). The amount of chemical A after 20 minutes is approximately 38.506 pounds.

Key Concepts

Chemical Solution Problems

Chemical solution problems are common in various scientific and engineering fields, particularly in chemistry and environmental engineering. These problems involve determining the concentration or amount of a particular chemical within a solution as it changes over time.
For instance:

• Mixing of Chemicals: Different chemicals are often mixed in a tank, where the rate of input and output can affect the concentration of each chemical in the mixture.
• Constant Volume: The exercise here describes a situation where the total volume of the solution remains constant, since the inflow and outflow rates are equal. This is a crucial premise as it simplifies the mathematical modelling.
• Problem Objective: Our primary goal in such problems is to find out how much of a specific chemical is left in the solution after a certain period, which depends on multiple factors like rate of inflow, rate of outflow, and initial concentration.

Understanding chemical solution problems are essential as they apply to real-world scenarios such as water treatment processes and pharmaceutical formulations.

Rate of Change

The rate of change is foundational in understanding differential equations, particularly in describing how quantities change over time. In the context of this exercise, it is the rate at which the amount of chemical changes in the tank.
In this scenario, the elements contributing to the rate of change include:

• Inflow Rate: The inflow rate of the chemical, calculated by multiplying the flow rate by the concentration of the incoming solution. Here it's 3 gallons per minute times 2 pounds per gallon, resulting in 6 pounds per minute.
• Outflow Rate: Determined by the concentration of the chemical in the tank and the outflow rate. If at some time the chemical's concentration is \( \frac{y(t)}{20} \), then the outflow is \( 3 \times \frac{y(t)}{20} \).
• Net Rate of Change: The differential equation \( \frac{dy}{dt} = 6 - \frac{3}{20} y \) represents the net rate, showing that the change in the amount of chemical is the difference between the in-rate and out-rate.

This net rate of change helps us understand the behavior of the chemical amount over time.

Integrating Factor Method

The integrating factor method is a powerful tool used to solve first-order linear differential equations. It facilitates transforming the original differential equation into an easily integrable form.
Here's how it applies:

• Linear Differential Equation: The exercise presents us with \( \frac{dy}{dt} + \frac{3}{20}y = 6 \), a first-order linear differential equation.
• Integrating Factor: The integrating factor \( \mu(t) \) is derived as \( e^{\int \frac{3}{20} dt} = e^{\frac{3}{20}t} \). This factor is multiplied across the equation to assist in integration.
• Transformation: Using the integrating factor, the equation becomes \( \frac{d}{dt}[y e^{\frac{3}{20}t}] = 6 e^{\frac{3}{20}t} \), simplifying further integration and solution derivation.

The method streamlines solving these equations, ultimately allowing us to determine the equation for chemical concentration over time.

Separable Equations

Separable equations are a special type of differential equations where variables can be separated on each side of the equation, making them straightforward to solve through direct integration.
In this exercise:

• Separation: Although the main method used is the integrating factor, separable equations can alternatively solve this problem as the equation can be rewritten such that terms involving \( y \) and terms involving \( t \) are on opposite sides.
• Integration: Once separated, each side of the differential equation is integrated to find a solution function \( y(t) \).
• Application: While the integrating factor method was more suitable here due to linear terms, recognizing the separability confirms the solution’s validity and provides more ways to approach similar problems.

Understanding separable equations broadens the tools available for tackling various differential equations.