Question

Initially, a 200-gallon tank is filled with pure water. At time \(t=0\), a salt concentration with 3 pounds of salt per gallon is added to the container at the rate of 4 gallons per minute, and the well-stirred mixture is drained from the container at the same rate. a. Find the number of pounds of salt in the container as a function of time. b. How many minutes does it take for the concentration to reach 2 pounds per gallon? c. What does the concentration in the container approach for large values of time? Does this agree with your intuition? d. Assuming that the tank holds much more than 200 gallons, and everything is the same except that the mixture is drained at 3 gallons per minute, what would the answers to parts a and b become?

Short answer

a. \(y(t) = 150 + 50e^{-t/50}\) pounds. b. Approximately 69.3 minutes. c. 0.75 pounds/gallon. d. \(y(t) = 600 - 400e^{-(3/200)t}\); Approximately 46.2 minutes.

Step-by-step solution

Setting up the differential equation

Let \(y(t)\) denote the amount of salt in pounds in the tank at time \(t\), then \(y'(t)\) is the rate of change of salt with respect to time. The rate of change of salt is the rate of salt coming in minus the rate of salt going out, so we have \[y'(t) = \text{rate in}-\text{rate out} = 4*3 - 4*(y(t)/200),\] because the rate in is the inflow rate (4 gal/min) times the concentration (3 lb/gal), and the rate out is the outflow rate (4 gal/min) times the concentration \(y(t)/200\) lb/gal.

Solving the differential equation

We simplify the differential equation by dividing through by 4: \[y'(t) = 3 - y(t)/50.\] Solve this first order linear differential equation using an integrating factor, we can obtain the solution: \[y(t) = 150 + 50e^{-t/50}.\] This represents the amount of salt in the tank at any time \(t\).

Find the time for concentration to reach 2 pounds per gallon

The concentration of salt in the tank is the amount of salt divided by the volume of liquid, that is, \(y(t)/200\). Set this equal to 2 and solve for \(t\) gives \(t = -50 ln(1/4) \approx 69.3\) minutes.

Calculating limit as time approaches infinity

By finding the limit of the function \(y(t)\) as \(t\) approaches infinity, we can find the concentration the tank approaches for large values of time. The limit is \(\lim\_{t \to \infty} y(t) = \lim\_{t \to \infty} (150 + 50e^{-t/50}) = 150\). The concentration is thus 150/200 or 0.75 lb/gal. This agrees intuitively since the inflow concentration, which is the ultimate concentration of the well-mixed solution, is less than the initial concentration of the pure water.

Adjust rate and repeat steps

If the mixture is drained at the rate of 3 gallons per minute instead of 4, the differential equation becomes \[y'(t) = 3 - y(t)/200,\] and the solution is \[y(t) = 600 - 400e^{-(3/200)t}.\] Solve for \(t\) when the concentration \(y(t)/200\) is equal to 2 gives \(t = -(200/3) ln(1/2) \approx 46.2\) minutes.

Key concepts

Salt Concentration Problem

Understanding the salt concentration problem in a dynamic system like a tank involves setting up a real-world scenario in the context of a differential equation. Imagine a tank that initially contains pure water. When a saline solution is added at a constant rate, and the mixed solution is drained at an equal rate, we must calculate how the salt concentration changes over time.

This situation is modeled mathematically by a differential equation which takes into account the rate of salt addition and the rate of salt removal. The equation reflects the balance between these two processes. As the solution is stirred well, the concentration of salt at any moment is uniform throughout the tank, simplifying the calculations. The primary goal is to find the function that represents the amount of salt in the tank at Time 't', and from there, derive the salt concentration.

Rate of Change

The rate of change is arguably one of the most central concepts in differential equations and physics, representing how a quantity changes over time. In our salt tank problem, the rate of change is modeled by the derivative of the salt amounts with respect to time, denoted as \(y'(t)\).

This derivative is found by subtracting the rate of salt leaving the tank from the rate of salt entering the tank. By understanding and finding the rate of change, we formulate a relationship that can be used to predict the future behavior of the system under the given conditions.

Integrating Factor

An integrating factor is a mathematical tool used to solve linear first-order differential equations. The purpose of an integrating factor is to multiply both sides of the differential equation by a function that enables the left-hand side to be written as the derivative of a product.

When we use an integrating factor in our salt tank problem, it simplifies the equation and makes it solvable. In our step by step solution, it transformed our equation into one where the amount of salt at time 't' can be determined directly. While the concept might seem daunting, it is merely a technique that makes otherwise complex differential equations more manageable and solvable.

Limit of a Function

The limit of a function as a variable approaches a specified input value (in our case, infinity) is a keystone in understanding long-term behavior in calculus. For differential equations modeling real-world scenarios, like the salt concentration in a tank, calculating the limit helps predict the eventual outcome given certain conditions.

In the salt concentration problem, finding the limit as time approaches infinity shows us the concentration level will stabilize at a certain value, after all dynamic changes have subsided. This is significant in both theoretical and practical applications as it gives us insight into the steady-state behavior of the system.