A Tank Has 150 Liters Of Water In It At Time { T=0 $}$ Hours. Water Begins To Be Pumped Into The Tank At Time { T=0 $}$. A Different Pipe Is Draining Water From The Tank Starting At { T=0 $}$.Water Is Being Removed From

Mar 10, 2025 by ADMIN 220 views

**A Tank Filling and Draining Problem: A Mathematical Analysis**

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 150 liters of water in it at time $t=0$ hours. Water begins to be pumped into the tank at a rate of 5 liters per hour starting at time $t=0$ . A different pipe is draining water from the tank at a rate of 2 liters per hour starting at time $t=0$ . We want to find the amount of water in the tank at any given time $t$ .

Mathematical Modeling

To solve this problem, we need to set up a differential equation that describes the rate of change of the amount of water in the tank. Let $V(t)$ be the amount of water in the tank at time $t$ . Then, the rate of change of $V(t)$ is given by the derivative $\frac{dV}{dt}$ .

The rate at which water is being pumped into the tank is 5 liters per hour, so the rate at which water is being added to the tank is $5$ liters per hour. The rate at which water is being drained from the tank is 2 liters per hour, so the rate at which water is being removed from the tank is $2$ liters per hour.

Therefore, the differential equation that describes the rate of change of the amount of water in the tank is:

\frac{dV}{dt} = 5 - 2 = 3

This equation says that the rate of change of the amount of water in the tank is 3 liters per hour.

Solving the Differential Equation

To solve this differential equation, we need to find the general solution of the equation. The general solution of a differential equation is a function that satisfies the equation.

In this case, the general solution of the differential equation is:

V(t) = 3t + C

where $C$ is a constant.

To find the value of $C$ , we need to use the initial condition. The initial condition is that the amount of water in the tank at time $t=0$ is 150 liters. Therefore, we have:

V(0) = 150 = 3(0) + C

Solving for $C$ , we get:

C = 150

Therefore, the particular solution of the differential equation is:

V(t) = 3t + 150

Interpretation of the Results

The solution $V(t) = 3t + 150$ tells us that the amount of water in the tank at time $t$ is given by the equation $V(t) = 3t + 150$ . This means that the amount of water in the tank is increasing at a rate of 3 liters per hour.

We can also use this equation to find the amount of water in the tank at any given time $t$ . For example, if we want to find the amount of water in the tank at time $t=2$ hours, we can plug in $t=2$ into the equation:

V(2) = 3(2) + 150 = 6 + 150 = 156

Therefore, the amount of water in the tank at time $t=2$ hours is 156 liters.

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. (2015). Differential equations and boundary value problems: Computing and modeling. Pearson.

Mathematical Derivations

Derivation of the Differential Equation

Let $V(t)$ be the amount of water in the tank at time $t$ . Then, the rate of change of $V(t)$ is given by the derivative $\frac{dV}{dt}$ .

Therefore, the differential equation that describes the rate of change of the amount of water in the tank is:

\frac{dV}{dt} = 5 - 2 = 3

Solution of the Differential Equation

To solve this differential equation, we need to find the general solution of the equation. The general solution of a differential equation is a function that satisfies the equation.

In this case, the general solution of the differential equation is:

V(t) = 3t + C

where $C$ is a constant.

To find the value of $C$ , we need to use the initial condition. The initial condition is that the amount of water in the tank at time $t=0$ is 150 liters. Therefore, we have:

V(0) = 150 = 3(0) + C

Solving for $C$ , we get:

C = 150

Therefore, the particular solution of the differential equation is:

V(t) = 3t + 150$<br/> **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&A

Q: What is the initial condition of the problem?

A: The initial condition of the problem is that the amount of water in the tank at time $t=0$ is 150 liters.

Q: What is the rate at which water is being pumped into the tank?

A: The rate at which water is being pumped into the tank is 5 liters per hour.

Q: What is the rate at which water is being drained from the tank?

A: The rate at which water is being drained from the tank is 2 liters per hour.

Q: What is the differential equation that describes the rate of change of the amount of water in the tank?

A: The differential equation that describes the rate of change of the amount of water in the tank is: