Problem 45 - Calculus Single Variable

Question

A \(1000 \mathrm{~L}\) tank initially contains a \(200 \mathrm{~L}\) solution in which \(24 \mathrm{~kg}\) of salt is dissolved. Beginning at time \(t=0,\) an inlet valve allows fresh water to flow into the tank at the constant rate of \(12 \mathrm{~L} / \mathrm{min},\) and an outlet valve is opened so that \(16 \mathrm{~L} / \mathrm{min}\) of the solution is drained. How much salt does the tank contain after 25 minutes?

Step-by-Step Solution

Answer

After 25 minutes, there is 1.5 kg of salt in the tank.

Step 1: Define Variables and Initial Conditions

Let \( S(t) \) be the amount of salt in the tank at time \( t \) minutes. Initially, \( S(0) = 24 \text{ kg} \) because there is 24 kg of salt in the tank. The initial volume of the solution in the tank is \( V_0 = 200 \text{ L} \). Fresh water is entering at \( 12 \text{ L/min} \) and the solution is being drained at \( 16 \text{ L/min} \). The volume \( V(t) \) at time \( t \) will be \( 200 - 4t \text{ L} \) since the tank loses \( 4 \text{ L/min} \).

Step 2: Set Up the Differential Equation

Since the salt only exits with the solution leaving the tank, the rate at which salt leaves the tank is proportional to the concentration of the salt in the tank. The rate of change of salt in the tank can be expressed as \( \frac{dS}{dt} = -16 \frac{S(t)}{V(t)} \). Substitute \( V(t) = 200 - 4t \): \( \frac{dS}{dt} = -16 \frac{S(t)}{200 - 4t} \).

Step 3: Solve the Differential Equation

This is a separable differential equation. Separate variables to get \( \frac{dS}{S} = -16 \frac{dt}{200 - 4t} \). Integrating both sides gives \( \ln |S| = -4 \ln |200 - 4t| + C_{1} \). Simplify to find \( S = C \cdot (200 - 4t)^{-4} \).

Step 4: Determine the Constant of Integration

Use the initial condition \( S(0) = 24\) to find \( C \). When \( t = 0\), \( S = 24 \) and \( V(0) = 200 \): \( 24 = C(200)^{-4} \). So, \( C = 24 \times 200^{4} \).

Step 5: Find the Amount of Salt after 25 Minutes

Substitute \( t = 25 \) into the equation for \( S(t) \): \( S(t) = 24 \times 200^{4} \times (200 - 4 \times 25)^{-4} \). Calculate \( S(25) = 24 \times \left( \frac{200}{100} \right)^{-4} = 24 \times 16 = \frac{24}{16} = 1.5 \text{ kg} \).

Key Concepts

Separable Differential Equations

Separable differential equations are a special class of differential equations where the variables can be separated on either side of the equation. This means you can rewrite the equation in a form where each variable appears on a separate side, making them much easier to solve.

In our exercise, we explore a situation involving a salt solution in a tank where fresh water enters while the salt solution is drained at different rates. The goal is to determine how much salt remains after a period of time. The mathematical model representing the rate of change of the salt can be expressed as \( \frac{dS}{dt} = -16 \frac{S(t)}{V(t)} \), where \( S(t) \) is the amount of salt at time \( t \), and \( V(t) \) is the volume of the solution at that time.

Since this is a separable equation, you can isolate variables to one side of the equation: \( \frac{dS}{S} = -16 \frac{dt}{200 - 4t} \). Both sides of the equation are then integrated separately, allowing us to obtain the formula for \( S(t) \). This flexibility in solving makes separable differential equations a favorite for scenarios where change over time needs to be analyzed.

Rate of Change

The rate of change is an essential concept in calculus, particularly in differential equations where it quantifies how a certain amount changes over time. In the context of our problem, it refers to how the quantity of salt changes with respect to time as fresh water enters and the salt solution drains.

At the core, the rate of change here is given by \( \frac{dS}{dt} \), which describes how the salt content in the tank decreases over time due to the outflow. This rate is directly proportional to the concentration of salt in the tank at any given time \( S(t)/V(t) \).

The actual rate at which the salt content drops is influenced by it's rate of outflow (16 L/min) and how it is initially mixed in the remaining solution. This establishes the basis for setting up the differential equation. Understanding the rate of change helps in predicting behaviors within systems dynamically changing over time.

Initial Conditions

Initial conditions serve as the foundation upon which particular solutions to differential equations are built. They establish the necessary information to solve for constants in the general solution of the differential equation, making them absolutely essential in finding unique solutions.

In our problem, the initial condition is that at time \( t = 0 \), the tank contains 24 kg of salt in a 200 L solution. Thus, \( S(0) = 24 \text{ kg} \). This particular piece of information allows us to solve for the integration constant \( C \) in the equation \( S = C \cdot (200 - 4t)^{-4} \).

Using the initial condition in our equation, \( 24 = C(200^{-4}) \), we can find \( C \) to precisely model the salt quantity at any given time. Initial conditions anchor the model to reality, ensuring solutions are relevant to the specific scenario.