Question

Two tanks are interconnected. Tank X initially contains 90 liters of brine in which there is dissolved \(3 \mathrm{~kg}\) of salt, and tank \(\mathrm{Y}\) initially contains 90 liters of brine in which there is dissolved \(2 \mathrm{~kg}\) of salt. Starting at time \(t=0,(1)\) pure water flows into tank \(\mathrm{X}\) at the rate of \(4.5\) liters/min, (2) brine flows from tank \(\mathrm{X}\) into tank \(\mathrm{Y}\) at the rate of \(6 \mathrm{liters} / \mathrm{min},(3)\) brine is pumped from tank \(\mathrm{Y}\) back into tank \(\mathrm{X}\) at the rate of \(1.5\) liters/min, and (4) brine flows out of tank \(Y\) and away from the system at the rate of \(4.5\) liters/min. The mixture in each tank is kept uniform by stirring. How much salt is in each tank at any time \(t>0 ?\)

Short answer

The amount of salt in each tank at any given time \(t>0\) can be found using the following expressions: - Tank X: \(x(t)=\frac{1}{4}(135e^{\frac{t}{30}}-e^{\frac{t}{10}}+90)\) - Tank Y: \(y(t)=\frac{1}{3}(45-15e^{\frac{t}{30}}+15e^{\frac{t}{10}}-30e^{\frac{t}{20}})\)

Step-by-step solution

Define Variables

Let \(x(t)\) be the amount of salt (in kg) in tank X at time \(t\), and let \(y(t)\) be the amount of salt (in kg) in tank Y at time \(t\).

Determine the rate of salt movement

We will now calculate the rate of salt entering or leaving each tank (in kg/min) based on the given flow rates (in liters/min): 1. Pure water enters tank X at a rate of 4.5 liters/min, so the salt content remains the same. 2. The salt leaves tank X at a rate of \(\frac{6}{90}x(t) \mathrm{~kg/min}\) and enters tank Y at the same rate. 3. The salt is pumped from tank Y to tank X at a rate of \(\frac{1.5}{90}y(t) \mathrm{~kg/min}\). 4. Salt flows out of tank Y at a rate of \(\frac{4.5}{90}y(t) \mathrm{~kg/min}\).

Set up differential equations

Using the rates calculated in the previous step, we can write the following system of differential equations: \[\frac{dx}{dt}=-\frac{6}{90}x(t)+\frac{1.5}{90}y(t)\] \[\frac{dy}{dt}=\frac{6}{90}x(t)-\frac{4.5}{90}y(t)\]

Solve the differential equations

To solve these differential equations, we can use the method of integrating factors. For the first equation, our integrating factor is \(e^{-\frac{1}{15}t}\). We then multiply the entire equation by this factor: \[e^{-\frac{1}{15}t}\frac{dx}{dt}-\frac{1}{15}e^{-\frac{1}{15}t}x(t)=\frac{1}{60}e^{-\frac{1}{15}t}y(t)\] Now, we notice that the left side of the equation is the derivative of \(e^{-\frac{1}{15}t}x(t)\). Using integration, we find: \[x(t)=\frac{1}{4}(135e^{\frac{t}{30}}-e^{\frac{t}{10}}+90)\] For the second equation, our integrating factor is \(e^{\frac{1}{20}t}\). We then multiply the entire equation by this factor: \[e^{\frac{1}{20}t}\frac{dy}{dt}+\frac{1}{20}e^{\frac{1}{20}t}y(t)=\frac{1}{15}e^{\frac{1}{20}t}x(t)\] Once again, we notice that the left side of the equation is the derivative of \(e^{\frac{1}{20}t}y(t)\). Using integration, we find: \[y(t)=\frac{1}{3}(45-15e^{\frac{t}{30}}+15e^{\frac{t}{10}}-30e^{\frac{t}{20}})\]

Interpret the solution

The expressions for the amount of salt in each tank at any given time t > 0 are as follows: - Tank X: \(x(t)=\frac{1}{4}(135e^{\frac{t}{30}}-e^{\frac{t}{10}}+90)\) - Tank Y: \(y(t)=\frac{1}{3}(45-15e^{\frac{t}{30}}+15e^{\frac{t}{10}}-30e^{\frac{t}{20}})\) These expressions give the amount of salt in each tank for any time \(t>0\).

Key concepts

Mathematical Modeling

Mathematical modeling is a fundamental tool used to describe real-world systems in terms of mathematical equations. It enables us to study the behavior of systems, predict outcomes, and test hypotheses in a controlled and replicable manner. In our exercise, we model the transfer of salt between two tanks using mathematical equations that represent the rate of change of salt concentration over time. This process begins with the identification of relevant variables - the amount of salt in each tank - and proceeds with the formulation of relationships based on flow rates and salt concentrations. The model takes into account the inflow and outflow of liquid and salt, ensuring that the equations accurately represent the physical setup. By carefully analyzing each step of the process, the model allows us to predict the salt concentration in each tank at any given time. To ensure clarity for students, it is essential to communicate the purpose and assumptions behind each variable and equation, thereby demystifying the modeling process and increasing its educational value.

In our example, the system's dynamics are captured by a set of ordinary differential equations (ODEs) that describe how the salt content in each tank changes over time. This forms the foundation of our mathematical model. For students, understanding the transition from the practical scenario - two interconnected tanks with fluid dynamics - to its mathematical representation is critical. It's important to emphasize that mathematical modeling is iterative and may require refinements based on observations or additional data. Used effectively, mathematical modeling in education not only enhances problem-solving skills but also provides a lens through which students can view complex systems in a structured and analytical way.

Systems of Differential Equations

When dealing with interconnected components or variables, such as the tanks in our exercise, a system of differential equations is often needed to describe the behavior of each variable concurrently. A differential equation is an equation that involves an unknown function and its derivatives, indicating how the function changes over time or other variables. When we have two or more such unknown functions, which depend on each other, we use systems of differential equations to describe their relations.

In the context of our tank problem, we have two functions: the amount of salt in tank X, denoted as \(x(t)\), and the amount of salt in tank Y, denoted as \(y(t)\). These amounts are not independent; the rate at which salt enters or leaves one tank will affect the amount of salt in the other. This interdependence requires us to solve the functions simultaneously because they evolve together over time. When explaining systems of differential equations to students, it is helpful to use visuals or analogies that emphasize the connected nature of the system's components. For example, explaining the concept using a flow diagram showing how water and salt move between the tanks can solidify their understanding. It is also beneficial to discuss how changes in one part of the system might impact the whole, highlighting the importance of considering all variables when analyzing complex systems.

Integrating Factors

An integrating factor is a mathematical tool used to solve certain types of differential equations, particularly linear first-order equations. It involves finding a function that, when multiplied by the original equation, transforms it into an equation with the derivative of a product on one side. This makes the equation easier to solve, as we can then integrate both sides with respect to the independent variable to find the solution.

In our tank exercise, we applied integrating factors to solve the linear first-order differential equations for \(x(t)\) and \(y(t)\). The correct integrating factor for each equation depends on the coefficients of \(x\) and \(y\). Multiplying the entire original differential equation by the integrating factor simplifies it by creating a derivative of a product, allowing us to integrate and solve for the unknown function. It's essential to clarify to students how the integrating factor is derived and why it's useful. An effective approach would be to work through a simpler example first, demonstrating each step in finding and applying the integrating factor, before diving into the more complex system of equations as seen in the tank problem. By understanding the concept of integrating factors, students will gain a powerful method for solving a broad class of differential equations, equipping them with the skills to tackle many problems in both mathematics and applied sciences.