Question

A tank originally contains 100 gal of fresh water. At time \(t=0\), a solution containing \(0.2 \mathrm{lb}\) of salt per gallon begins to flow into the tank at a rate of \(3 \mathrm{gal} / \mathrm{min}\) and the well-stirred mixture flows out of the tank at the same rate. (a) How much salt is in the tank after \(10 \mathrm{~min}\) ? (b) Does the amount of salt approach a limiting value as time increases? If so, what is this limiting value and what is the limiting concentration?

Short answer

Based on the solution provided, the amount of salt in the tank after 10 minutes is approximately 22.31 pounds and there is a limiting amount of salt in the tank at 100 pounds, with a limiting concentration of 1 pound per gallon.

Step-by-step solution

Define Variables and Setup the Equation

Let \(Q(t)\) represent the amount of salt in the tank (in pounds) at time \(t\) (in minutes). The rate of change of salt in the tank, \(dQ/dt\), is equal to the salt coming in minus the salt going out. The salt is coming in at a rate of \(3\) gallons per minute, with a concentration of \(0.2\) pounds of salt per gallon. Therefore, the rate of salt coming in is: $$ \frac{dQ_{in}}{dt} = 3 \mathrm{~gal/min} \times 0.2 \mathrm{~lb/gal} = 0.6 \mathrm{~lb/min}. $$ Since the well-stirred mixture flows out of the tank at the same rate, the concentration of salt in the tank is equal to \(Q(t) / 100\). Consequently, the rate of salt going out is: $$ \frac{dQ_{out}}{dt} = 3 \mathrm{~gal/min} \times \frac{Q(t)}{100 \mathrm{~gal}}. $$ Now, we can write the equation for the rate of change of salt in the tank: $$ \frac{dQ}{dt} = \frac{dQ_{in}}{dt} - \frac{dQ_{out}}{dt} = 0.6 - \frac{3Q(t)}{100}. $$

Solve the Differential Equation

We will now solve the differential equation for \(Q(t)\). This is a first-order linear differential equation, and we can solve it using an integrating factor. Let \(μ(t) = e^{\int -\frac{3}{100} dt} = e^{-\frac{3}{100}t}\) be the integrating factor. Multiply both sides of the equation by the integrating factor: $$ e^{-\frac{3}{100}t} \frac{dQ}{dt} + \frac{3}{100}e^{-\frac{3}{100}t} Q(t) = 0.6 e^{-\frac{3}{100}t}. $$ The left side of the equation is now the derivative of the product \((Q(t) e^{-\frac{3}{100}t})\) with respect to \(t\). Thus, we can rewrite the equation as: $$ \frac{d}{dt}(Q(t) e^{-\frac{3}{100}t}) = 0.6 e^{-\frac{3}{100}t}. $$ Now, integrate both sides with respect to \(t\): $$ Q(t)e^{-\frac{3}{100}t} = 20 - 20e^{-\frac{3}{100}t} + C. $$ At \(t = 0\), the tank contains only fresh water, so \(Q(0) = 0\). Therefore, \(C = -20\). Then, the function for the amount of salt in the tank is: $$ Q(t) = 100 - 100 e^{-\frac{3}{100}t}. $$

Calculate the Amount of Salt After 10 Minutes

Evaluate the function \(Q(t)\) at \(t = 10\) to determine the amount of salt in the tank after \(10\) minutes: $$ Q(10) = 100 - 100 e^{-\frac{3}{100}(10)} \approx 22.31 \mathrm{~lb}. $$ So, there are approximately \(22.31\) pounds of salt in the tank after \(10\) minutes.

Determine the Limiting Amount of Salt

To find the limiting amount of salt, calculate the limit of \(Q(t)\) as \(t\) approaches infinity: $$ \lim_{t \to \infty} Q(t) = \lim_{t \to \infty} (100 - 100 e^{-\frac{3}{100}t}) = 100. $$ The limiting amount of salt in the tank is \(100\) pounds. The limiting concentration of the salt in the tank can be found by dividing the limiting amount of salt by the constant volume of the tank (100 gallons): $$ \mathrm{Limiting~concentration} = \frac{100 \mathrm{~lb}}{100 \mathrm{~gal}} = 1 \mathrm{~lb/gal}. $$ As time increases, the amount of salt in the tank approaches a limiting value of \(100\) pounds, and the limiting concentration is \(1\) pound per gallon.

Key concepts

First-order linear differential equation

A first-order linear differential equation is among the fundamental concepts encountered in calculus and is essential for solving many real-world problems, including mixing problems.

In simple terms, these equations involve rates of change (represented by derivatives) that are proportional to the current value of the function. The general form is often written as
\[\frac{{dQ}}{{dt}} + P(t)Q = G(t)\]
where \(Q\) represents the quantity we're interested in, \(t\) is time, \(P(t)\) a function of time which modifies the rate of change, and \(G(t)\) represents an external input or source term. In the context of our tank-mixing problem, \(Q(t)\) is the amount of salt, \(P(t)\) is the rate at which the tank's contents are diluted, and \(G(t)\) is the rate at which salt enters the tank.

Understanding these equations requires recognizing patterns and discerning how changes in one variable affect the other. When teaching this topic, it's helpful to stress the importance of setting up the equation correctly and ensuring that all units align, which is a practical analytical skill for students.

Integrating factor method

The integrating factor method is a powerful technique used to solve first-order linear differential equations. This approach involves finding a function, called the integrating factor, that when multiplied by the original equation, simplifies it to a format that's easy to integrate.

The general formula for the integrating factor, \(\mu(t)\), is derived from the original differential equation and is often expressed as
\[\mu(t) = e^{\int P(t) dt}\]
The magic of this method is that it takes advantage of certain properties of derivatives and exponential functions, transforming our problem into one where we can directly integrate both sides to find the solution. In the context of our mixing problem involving a tank with an inflow and outflow of liquid, applying the integrating factor eliminates the need to chase the effects of continuously varying concentrations through time. Instead, it allows for a straightforward calculation of the amount of salt in the tank at any given time. Students often struggle with this step, so emphasizing the method's reliance on the fundamental properties of exponents and derivatives can aid comprehension.

Limiting value in mixing problems

In mixing problems, the 'limiting value' is a key concept that describes the final steady-state or equilibrium condition that the system approaches over time. Knowing how to calculate the limiting value is crucial since it represents the ultimate behavior of the system as affected by the rates of input and output.

Given enough time, and assuming a constant process, a mixing system will balance the inputs and outputs resulting in a uniform concentration that no longer changes — this is the limiting value. Mathematically, we find this by taking the limit of our salt quantity function \(Q(t)\) as time \(t\) goes to infinity:
\[\lim_{{t \to \infty}} Q(t)\]
For instance, in our tank scenario, we calculated this limit to find out that after an infinitely long time, the amount of salt would stabilize at 100 pounds. Understanding this concept is critical when considering real-world applications, like environmental engineering, where limiting values can represent the concentration of a pollutant in a body of water or a reactor at equilibrium. Students should be guided through the relevant calculus concepts, ensuring they recognize the implications of the equilibrium in practical scenarios.

Modeling with differential equations

Modeling with differential equations involves translating real-world scenarios into mathematical terms to predict behavior and outcomes. This approach is indispensable in science and engineering, allowing us to understand dynamic systems in a precise and quantitative way.

In our example of the mixing problem, we created a model that describes how salt enters and leaves a tank. By expressing the changes in salt concentration with a differential equation, we gain a tool that can predict the amount of salt at any given moment. This process, common in engineering and physics, relies on identifying rates of change and expressing them in terms of derivatives. It combines observational data (like the flow rate in and out of the tank and the concentration of the inflow) with mathematical structures to represent the phenomena at play.

When teaching this concept, it's essential to focus on real-world interpretation of the rates involved in the model and encourage students to visualize the physical process. This helps bridge the gap between abstract equations and tangible reality, improving their understanding not only of the mathematics but also of the system being modeled.