A Tank Has 50 Liters Of Water In It At Time T = 0 T = 0 T = 0 Minutes. Water Begins To Be Pumped Into The Tank At Time T = 0 T = 0 T = 0 . A Different Pipe Is Draining Water From The Tank Starting At T = 0 T = 0 T = 0 .Water Is Being Pumped Into The Tank At

Introduction

In this article, we will delve into a classic problem in mathematics involving a tank that is being filled and drained simultaneously. We will analyze the situation, derive the necessary equations, and solve for the amount of water in the tank at any given time. This problem is a great example of how mathematical modeling can be used to understand and solve real-world problems.

The Problem

A tank has 50 liters of water in it at time $t = 0$ minutes. Water begins to be pumped into the tank at a rate of $r_1$ liters per minute, starting at time $t = 0$. At the same time, a different pipe is draining water from the tank at a rate of $r_2$ liters per minute. We want to find the amount of water in the tank at any given time $t$.

Mathematical Modeling

Let $V(t)$ be the amount of water in the tank at time $t$. We can model the situation using the following differential equation:

$$\frac{dV}{dt} = r_1 - r_2$$

This equation states that the rate of change of the amount of water in the tank is equal to the rate at which water is being pumped in minus the rate at which water is being drained.

Initial Condition

We are given that the tank initially contains 50 liters of water, so we have the initial condition:

$$V(0) = 50$$

Solving the Differential Equation

To solve the differential equation, we can use the method of separation of variables. We can rewrite the equation as:

$$\frac{dV}{r_1 - r_2} = dt$$

Integrating both sides, we get:

$$\int \frac{dV}{r_1 - r_2} = \int dt$$

Evaluating the integrals, we get:

$$\frac{V}{r_1 - r_2} = t + C$$

where $C$ is the constant of integration.

Applying the Initial Condition

We can apply the initial condition to find the value of $C$. Substituting $V(0) = 50$, we get:

$$\frac{50}{r_1 - r_2} = 0 + C$$

Solving for $C$, we get:

$$C = \frac{50}{r_1 - r_2}$$

Finding the General Solution

Substituting the value of $C$ back into the equation, we get:

$$\frac{V}{r_1 - r_2} = t + \frac{50}{r_1 - r_2}$$

Multiplying both sides by $r_1 - r_2$, we get:

$$V = (r_1 - r_2)t + 50$$

This is the general solution to the differential equation.

Interpretation

The general solution tells us that the amount of water in the tank at any given time $t$ is equal to the rate at which water is being pumped in minus the rate at which water is being drained, multiplied by the time, plus the initial amount of water in the tank.

Example

Suppose that water is being pumped into the tank at a rate of 2 liters per minute, and water is being drained from the tank at a rate of 1 liter per minute. We can use the general solution to find the amount of water in the tank at any given time $t$.

Substituting $r_1 = 2$ and $r_2 = 1$ into the general solution, we get:

$$V = (2 - 1)t + 50$$

Simplifying, we get:

$$V = t + 50$$

This means that the amount of water in the tank at any given time $t$ is equal to the time plus 50.

Conclusion

In this article, we analyzed a classic problem in mathematics involving a tank that is being filled and drained simultaneously. We derived the necessary equations, solved for the amount of water in the tank at any given time, and interpreted the results. This problem is a great example of how mathematical modeling can be used to understand and solve real-world problems.

References

• [1] Boyce, W. E., & DiPrima, R. C. (2012). Elementary differential equations and boundary value problems. John Wiley & Sons.
• [2] Edwards, C. H., & Penney, D. E. (2010). Differential equations and boundary value problems: Computing and modeling. Pearson Prentice Hall.

Glossary

• Differential equation: An equation that involves an unknown function and its derivatives.
• Separation of variables: A method of solving differential equations by separating the variables and integrating both sides.
• Initial condition: A condition that is specified at the beginning of a problem, such as the initial amount of water in the tank.
• General solution: A solution to a differential equation that is valid for all values of the independent variable.
• Particular solution: A solution to a differential equation that is valid for a specific value of the independent variable.
  A Tank Filling and Draining Problem: A Mathematical Analysis - Q&A ====================================================================

Introduction

In our previous article, we analyzed a classic problem in mathematics involving a tank that is being filled and drained simultaneously. We derived the necessary equations, solved for the amount of water in the tank at any given time, and interpreted the results. In this article, we will answer some of the most frequently asked questions about this problem.

Q: What is the initial condition for this problem?

A: The initial condition for this problem is that the tank initially contains 50 liters of water.

Q: What are the rates at which water is being pumped in and drained from the tank?

A: The rate at which water is being pumped in is $r_1$ liters per minute, and the rate at which water is being drained from the tank is $r_2$ liters per minute.

Q: How do we model this situation mathematically?

A: We can model this situation using the following differential equation:

$$\frac{dV}{dt} = r_1 - r_2$$

This equation states that the rate of change of the amount of water in the tank is equal to the rate at which water is being pumped in minus the rate at which water is being drained.

Q: How do we solve this differential equation?

A: We can solve this differential equation using the method of separation of variables. We can rewrite the equation as:

$$\frac{dV}{r_1 - r_2} = dt$$

Integrating both sides, we get:

$$\int \frac{dV}{r_1 - r_2} = \int dt$$

Evaluating the integrals, we get:

$$\frac{V}{r_1 - r_2} = t + C$$

where $C$ is the constant of integration.

Q: What is the general solution to this differential equation?

A: The general solution to this differential equation is:

$$V = (r_1 - r_2)t + 50$$

This equation tells us that the amount of water in the tank at any given time $t$ is equal to the rate at which water is being pumped in minus the rate at which water is being drained, multiplied by the time, plus the initial amount of water in the tank.

Q: What is the particular solution to this differential equation?

A: The particular solution to this differential equation is:

$$V = t + 50$$

This equation tells us that the amount of water in the tank at any given time $t$ is equal to the time plus 50.

Q: How do we interpret the results of this problem?

A: The results of this problem tell us that the amount of water in the tank at any given time $t$ is equal to the rate at which water is being pumped in minus the rate at which water is being drained, multiplied by the time, plus the initial amount of water in the tank.

Q: What are some real-world applications of this problem?

A: Some real-world applications of this problem include:

• Modeling the flow of water in a tank
• Modeling the flow of fluids in a pipe
• Modeling the growth of a population
• Modeling the spread of a disease

Q: What are some common mistakes to avoid when solving this problem?

A: Some common mistakes to avoid when solving this problem include:

• Not specifying the initial condition
• Not using the correct method of separation of variables
• Not evaluating the integrals correctly
• Not interpreting the results correctly

Conclusion

In this article, we answered some of the most frequently asked questions about the tank filling and draining problem. We hope that this article has been helpful in clarifying any confusion and providing a better understanding of this problem.

References

• [1] Boyce, W. E., & DiPrima, R. C. (2012). Elementary differential equations and boundary value problems. John Wiley & Sons.
• [2] Edwards, C. H., & Penney, D. E. (2010). Differential equations and boundary value problems: Computing and modeling. Pearson Prentice Hall.

Glossary

• Differential equation: An equation that involves an unknown function and its derivatives.
• Separation of variables: A method of solving differential equations by separating the variables and integrating both sides.
• Initial condition: A condition that is specified at the beginning of a problem, such as the initial amount of water in the tank.
• General solution: A solution to a differential equation that is valid for all values of the independent variable.
• Particular solution: A solution to a differential equation that is valid for a specific value of the independent variable.