Two Tanks Contain Water. One Tank Already Has 39 Gallons Of Water When 3 Gallons Per Minute Are Added. The Second Tank Has 95 Gallons Of Water When It Starts Leaking 5 Gallons Per Minute. If $m$ Represents Time In Minutes And $g$

Introduction

In this article, we will delve into a mathematical problem involving two tanks containing water. One tank already has a significant amount of water, while the other tank has a substantial amount of water but starts leaking at a certain rate. We will use mathematical equations to model the situation and analyze the changes in the water levels of both tanks over time.

Tank 1: Filling Up

The first tank already has 39 gallons of water. At a rate of 3 gallons per minute, water is being added to the tank. We can represent this situation using the equation:

$$\frac{dW}{dt} = 3$$

where $W$ is the amount of water in the tank at time $t$.

To find the amount of water in the tank at any given time, we can integrate the equation with respect to time:

$$W(t) = \int 3 dt = 3t + C$$

where $C$ is the constant of integration. Since the tank already has 39 gallons of water, we can set up the initial condition:

$$W(0) = 39$$

Substituting this into the equation, we get:

$$39 = 3(0) + C$$

Solving for $C$, we get:

$$C = 39$$

Therefore, the amount of water in the tank at any given time is:

$$W(t) = 3t + 39$$

Tank 2: Leaking Water

The second tank has 95 gallons of water and starts leaking at a rate of 5 gallons per minute. We can represent this situation using the equation:

$$\frac{dW}{dt} = -5$$

where $W$ is the amount of water in the tank at time $t$.

To find the amount of water in the tank at any given time, we can integrate the equation with respect to time:

$$W(t) = \int -5 dt = -5t + C$$

where $C$ is the constant of integration. Since the tank already has 95 gallons of water, we can set up the initial condition:

$$W(0) = 95$$

Substituting this into the equation, we get:

$$95 = -5(0) + C$$

Solving for $C$, we get:

$$C = 95$$

Therefore, the amount of water in the tank at any given time is:

$$W(t) = -5t + 95$$

Comparing the Two Tanks

Now that we have the equations for the amount of water in each tank, we can compare the two tanks. We can set up a system of equations to represent the situation:

$$W_1(t) = 3t + 39$$

$$W_2(t) = -5t + 95$$

where $W_1(t)$ is the amount of water in the first tank and $W_2(t)$ is the amount of water in the second tank.

We can solve this system of equations to find the time at which the two tanks have the same amount of water. To do this, we can set up the equation:

$$3t + 39 = -5t + 95$$

Solving for $t$, we get:

$$8t = 56$$

$$t = 7$$

Therefore, the two tanks have the same amount of water after 7 minutes.

Conclusion

In this article, we analyzed the situation of two tanks containing water. One tank was filling up at a rate of 3 gallons per minute, while the other tank was leaking at a rate of 5 gallons per minute. We used mathematical equations to model the situation and found that the two tanks have the same amount of water after 7 minutes.

Mathematical Derivations

Derivation of the Equation for Tank 1

The equation for the amount of water in the first tank is:

$$W(t) = 3t + 39$$

To derive this equation, we can use the following steps:

1. The rate at which water is being added to the tank is 3 gallons per minute.

2. We can represent this situation using the equation:

   $$\frac{dW}{dt} = 3$$

3. To find the amount of water in the tank at any given time, we can integrate the equation with respect to time:

   $$W(t) = \int 3 dt = 3t + C$$

4. Since the tank already has 39 gallons of water, we can set up the initial condition:

   $$W(0) = 39$$

5. Substituting this into the equation, we get:

   $$39 = 3(0) + C$$

6. Solving for $C$, we get:

   $$C = 39$$

7. Therefore, the amount of water in the tank at any given time is:

   $$W(t) = 3t + 39$$

Derivation of the Equation for Tank 2

The equation for the amount of water in the second tank is:

$$W(t) = -5t + 95$$

To derive this equation, we can use the following steps:

1. The rate at which water is leaking from the tank is 5 gallons per minute.

2. We can represent this situation using the equation:

   $$\frac{dW}{dt} = -5$$

3. To find the amount of water in the tank at any given time, we can integrate the equation with respect to time:

   $$W(t) = \int -5 dt = -5t + C$$

4. Since the tank already has 95 gallons of water, we can set up the initial condition:

   $$W(0) = 95$$

5. Substituting this into the equation, we get:

   $$95 = -5(0) + C$$

6. Solving for $C$, we get:

   $$C = 95$$

7. Therefore, the amount of water in the tank at any given time is:

   $$W(t) = -5t + 95$$

Solution to the System of Equations

We can solve the system of equations to find the time at which the two tanks have the same amount of water. To do this, we can set up the equation:

$$3t + 39 = -5t + 95$$

Solving for $t$, we get:

$$8t = 56$$

$$t = 7$$

Therefore, the two tanks have the same amount of water after 7 minutes.

References

• [1] Calculus: Early Transcendentals, James Stewart, 8th edition.
• [2] Differential Equations and Dynamical Systems, Lawrence Perko, 3rd edition.

Glossary

• Tank 1: The first tank, which is filling up at a rate of 3 gallons per minute.
• Tank 2: The second tank, which is leaking at a rate of 5 gallons per minute.
• Water level: The amount of water in a tank at a given time.
• Rate of change: The rate at which the water level in a tank is changing.
• System of equations: A set of equations that are used to solve a problem.
• Integration: The process of finding the antiderivative of a function.
• Derivative: The rate of change of a function with respect to its input.
  Two Tanks Containing Water: A Mathematical Analysis - Q&A ===========================================================

Introduction

In our previous article, we analyzed the situation of two tanks containing water. One tank was filling up at a rate of 3 gallons per minute, while the other tank was leaking at a rate of 5 gallons per minute. We used mathematical equations to model the situation and found that the two tanks have the same amount of water after 7 minutes. In this article, we will answer some frequently asked questions related to this problem.

Q&A

Q: What is the initial amount of water in Tank 1?

A: The initial amount of water in Tank 1 is 39 gallons.

Q: What is the rate at which water is being added to Tank 1?

A: The rate at which water is being added to Tank 1 is 3 gallons per minute.

Q: What is the initial amount of water in Tank 2?

A: The initial amount of water in Tank 2 is 95 gallons.

Q: What is the rate at which water is leaking from Tank 2?

A: The rate at which water is leaking from Tank 2 is 5 gallons per minute.

Q: How long does it take for the two tanks to have the same amount of water?

A: It takes 7 minutes for the two tanks to have the same amount of water.

Q: What is the amount of water in Tank 1 after 7 minutes?

A: The amount of water in Tank 1 after 7 minutes is 3(7) + 39 = 78 gallons.

Q: What is the amount of water in Tank 2 after 7 minutes?

A: The amount of water in Tank 2 after 7 minutes is -5(7) + 95 = 0 gallons.

Q: Why do the two tanks have the same amount of water after 7 minutes?

A: The two tanks have the same amount of water after 7 minutes because the rate at which water is being added to Tank 1 is equal to the rate at which water is leaking from Tank 2.

Q: What is the significance of this problem?

A: This problem is significant because it illustrates the concept of rates of change and how they can be used to model real-world situations.

Conclusion

In this article, we answered some frequently asked questions related to the problem of two tanks containing water. We hope that this article has provided a better understanding of the problem and its significance.

Mathematical Derivations

Derivation of the Equation for Tank 1

The equation for the amount of water in the first tank is:

$$W(t) = 3t + 39$$

To derive this equation, we can use the following steps:

1. The rate at which water is being added to the tank is 3 gallons per minute.

2. We can represent this situation using the equation:

   $$\frac{dW}{dt} = 3$$

3. To find the amount of water in the tank at any given time, we can integrate the equation with respect to time:

   $$W(t) = \int 3 dt = 3t + C$$

4. Since the tank already has 39 gallons of water, we can set up the initial condition:

   $$W(0) = 39$$

5. Substituting this into the equation, we get:

   $$39 = 3(0) + C$$

6. Solving for $C$, we get:

   $$C = 39$$

7. Therefore, the amount of water in the tank at any given time is:

   $$W(t) = 3t + 39$$

Derivation of the Equation for Tank 2

The equation for the amount of water in the second tank is:

$$W(t) = -5t + 95$$

To derive this equation, we can use the following steps:

1. The rate at which water is leaking from the tank is 5 gallons per minute.

2. We can represent this situation using the equation:

   $$\frac{dW}{dt} = -5$$

3. To find the amount of water in the tank at any given time, we can integrate the equation with respect to time:

   $$W(t) = \int -5 dt = -5t + C$$

4. Since the tank already has 95 gallons of water, we can set up the initial condition:

   $$W(0) = 95$$

5. Substituting this into the equation, we get:

   $$95 = -5(0) + C$$

6. Solving for $C$, we get:

   $$C = 95$$

7. Therefore, the amount of water in the tank at any given time is:

   $$W(t) = -5t + 95$$

Solution to the System of Equations

We can solve the system of equations to find the time at which the two tanks have the same amount of water. To do this, we can set up the equation:

$$3t + 39 = -5t + 95$$

Solving for $t$, we get:

$$8t = 56$$

$$t = 7$$

Therefore, the two tanks have the same amount of water after 7 minutes.

References

• [1] Calculus: Early Transcendentals, James Stewart, 8th edition.
• [2] Differential Equations and Dynamical Systems, Lawrence Perko, 3rd edition.

Glossary

• Tank 1: The first tank, which is filling up at a rate of 3 gallons per minute.
• Tank 2: The second tank, which is leaking at a rate of 5 gallons per minute.
• Water level: The amount of water in a tank at a given time.
• Rate of change: The rate at which the water level in a tank is changing.
• System of equations: A set of equations that are used to solve a problem.
• Integration: The process of finding the antiderivative of a function.
• Derivative: The rate of change of a function with respect to its input.