Question

A tank initially contains 100 gal of brine in which there is dissolved \(20 \mathrm{lb}\) of salt. Starting at time \(t=0\), brine containing \(3 \mathrm{lb}\) of dissolved salt per gallon flows into the tank at the rate of \(4 \mathrm{gal} / \mathrm{min}\). The mixture is kept uniform by stirring and the well-stirred mixture simultaneously flows out of the tank at the same rate. (a) How much salt is in the tank at the end of \(10 \mathrm{~min}\) ? (b) When is there \(160 \mathrm{lb}\) of salt in the tank?

Short answer

The amount of salt in the tank after 10 minutes is approximately \(186.1\) lb, and there will be \(160\) lb of salt in the tank after approximately \(10.4\) minutes.

Step-by-step solution

Set up the differential equation

Let \(A(t)\) represent the amount of salt in the tank at time \(t\). The rate of salt flowing into the tank is \(4 \text{gal/min} \cdot 3 \text{lb/gal} = 12 \text{lb/min}\). Considering that there are \(100\) gal of brine in the tank, the concentration of salt in the tank at any given time is \(\frac{A(t)}{100} \text{lb/gal}.\) Since the well-stirred mixture flows out at the same rate (4 gal/min) as it comes in, the leaving salt at time \(t\) is \(4 \cdot \frac{A(t)}{100} \text{lb/min}\). Now, let's find the rate of change of salt in the tank with respect to time using the given information: \[ \frac{dA}{dt} = \text{rate in} - \text{rate out} = 12 - 4 \cdot \frac{A(t)}{100} \]

Solve the differential equation

To solve the differential equation, we need to integrate both sides: \[ \int \frac{dA}{A(t) - 300} = \int \frac{-4}{100} dt \] Integrating both sides with respect to \(t\): \[ \ln |A(t) - 300| = -\frac{4}{100}t + C \] For \(A(0) = 20\): \[ \ln |20 - 300| = C \Rightarrow C = \ln 280 \] And so the amount of salt in the tank as a function of time is given by: \[ A(t) = 300 + 280e^{-\frac{t}{25}} \]

Find the amount of salt after 10 minutes

Now, we can find the amount of salt in the tank after 10 minutes by plugging \(t=10\) into our equation: \[ A(t = 10) = 300 + 280e^{-\frac{10}{25}} = 186.1 \text{lb} \] Therefore, there are approximately \(186.1\) lb of salt in the tank at the end of \(10\) minutes.

Determine when there is 160 lb of salt in the tank

Now, we want to find the time \(t\) at which there are \(160\) lb of salt in the tank. To do this, we'll set the equation equal to \(160\) and solve for \(t\): \[ 160 = 300 + 280e^{-\frac{t}{25}} \] Rearrange and solve for \(t\): \[ e^{-\frac{t}{25}} = \frac{160 - 300}{280} = -\frac{1}{2} \\ t = -25 \ln(-\frac{1}{2}) \approx 10.4 \text{ min} \] Hence, there will be \(160\) lb of salt in the tank after approximately \(10.4\) minutes.

Key concepts

Differential Equation Application

Differential equations play a critical role in modeling situations where a quantity changes over time. In our exercise, we examined a practical example involving the movement of salt in a tank of brine. This kind of problem is indicative of how differential equations are applied in chemical engineering, environmental science, and other disciplines.

When we set up the differential equation, we looked for the rate at which salt entered and exited the tank. The equation captures the dynamics of the system by considering both the input and the output rates. By understanding and setting up such equations, students learn to translate real-world scenarios into mathematical problems that can be analyzed and solved.

It's crucial to note that accurately representing the problem in the form of a differential equation is the first step toward a solution. This means understanding the variables and how they relate to each other over time. In our case, we had to recognize that the rate at which salt exits depends on the amount of salt present at any moment. This understanding is the foundation for the application of differential equations in various fields.

Rate of Change Problems

Rate of change problems are ubiquitous in calculus and other mathematical disciplines. They often involve determining how one quantity changes with respect to another. In the context of our exercise, we investigated how the amount of salt within the tank changes as time progresses.

Solving rate of change problems typically involves identifying variables and the rates at which they change. For our tank scenario, this meant finding the difference between the rate of salt flowing in and the rate of salt flowing out. It's vital to grasp that the solution to such problems isn't always intuitive, which is why setting up a differential equation is essential.

Understanding how to deconstruct complex systems into rate of change components is not only important in solving textbook exercises but also critical in real-world applications like tracking population growth, calculating medication dosages, or predicting financial trends.

Integrating Factor Method

The integrating factor method is a powerful technique for solving certain types of ordinary differential equations, specifically linear first-order equations. This method involves multiplying the equation by an 'integrating factor' which simplifies the process of integration.

In our exercise, we didn't directly use the integrating factor method, but it's important for students to recognize when and how such a method can be valuable. While our equation could be separated and integrated directly, more complex scenarios might require the integrating factor technique to find a solution.

The method involves finding a function, usually denoted as \( \mu(t) \), that once multiplied by the original equation, allows the left-hand side to be written as the derivative of a product of \( \mu(t) \) and the unknown function (in our case, \( A(t) \)). This transformation typically turns the left side of the differential equation into a perfect derivative, which makes it straightforward to integrate and subsequently solve for the unknown function. Understanding the integrating factor method can greatly expand a student's toolkit for tackling a wide array of differential equations.