A 28,000-gallon Swimming Pool Is Being Drained Using A Pump That Empties 700 Gallons Per Hour. Which Equation Models This Situation If $g$ Is The Number Of Gallons Remaining In The Pool And $t$ Is The Amount Of Time In Hours The

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A 28,000-gallon Swimming Pool Draining Situation: Modeling with Mathematics

In this article, we will explore a real-world scenario involving a 28,000-gallon swimming pool being drained using a pump that empties 700 gallons per hour. We will use mathematical equations to model this situation, focusing on the relationship between the number of gallons remaining in the pool and the amount of time the pump has been operating.

The problem states that a 28,000-gallon swimming pool is being drained at a rate of 700 gallons per hour. We are asked to find an equation that models this situation, where $g$ represents the number of gallons remaining in the pool and $t$ represents the amount of time in hours the pump has been operating.

To model this situation, we can use the concept of rate of change. The rate at which the pool is being drained is 700 gallons per hour, which means that the number of gallons remaining in the pool decreases by 700 gallons every hour. We can represent this relationship using the equation:

dgdt=−700\frac{dg}{dt} = -700

where $\frac{dg}{dt}$ represents the rate of change of the number of gallons remaining in the pool with respect to time.

To find the equation that models the situation, we need to solve the differential equation:

dgdt=−700\frac{dg}{dt} = -700

We can solve this equation by integrating both sides with respect to time:

∫dgdtdt=∫−700dt\int \frac{dg}{dt} dt = \int -700 dt

g(t)=−700t+Cg(t) = -700t + C

where $C$ is the constant of integration.

We know that initially, the pool contains 28,000 gallons of water. We can use this information to find the value of the constant $C$:

g(0)=28,000g(0) = 28,000

Substituting $t = 0$ into the equation, we get:

g(0)=−700(0)+Cg(0) = -700(0) + C

C=28,000C = 28,000

Now that we have found the value of the constant $C$, we can substitute it back into the equation:

g(t)=−700t+28,000g(t) = -700t + 28,000

This is the equation that models the situation, where $g(t)$ represents the number of gallons remaining in the pool after $t$ hours.

The equation $g(t) = -700t + 28,000$ represents the relationship between the number of gallons remaining in the pool and the amount of time the pump has been operating. We can use this equation to find the number of gallons remaining in the pool at any given time.

For example, if the pump has been operating for 4 hours, we can substitute $t = 4$ into the equation:

g(4)=−700(4)+28,000g(4) = -700(4) + 28,000

g(4)=−2,800+28,000g(4) = -2,800 + 28,000

g(4)=25,200g(4) = 25,200

This means that after 4 hours, there will be 25,200 gallons of water remaining in the pool.

In this article, we used mathematical equations to model a real-world scenario involving a 28,000-gallon swimming pool being drained using a pump that empties 700 gallons per hour. We found the equation that models this situation, where $g(t)$ represents the number of gallons remaining in the pool after $t$ hours. We can use this equation to find the number of gallons remaining in the pool at any given time, providing valuable insights into the draining process.

  • [1] Calculus: Early Transcendentals, 8th edition, by James Stewart
  • [2] Differential Equations and Dynamical Systems, 4th edition, by Lawrence Perko

For more information on mathematical modeling and differential equations, we recommend the following resources:

  • [1] Khan Academy: Differential Equations
  • [2] MIT OpenCourseWare: Differential Equations
  • [3] Wolfram MathWorld: Differential Equations
    A 28,000-gallon Swimming Pool Draining Situation: Modeling with Mathematics - Q&A

In our previous article, we explored a real-world scenario involving a 28,000-gallon swimming pool being drained using a pump that empties 700 gallons per hour. We used mathematical equations to model this situation, focusing on the relationship between the number of gallons remaining in the pool and the amount of time the pump has been operating.

In this article, we will answer some of the most frequently asked questions related to this scenario, providing additional insights and clarifications.

Q: What is the rate at which the pool is being drained?

A: The pool is being drained at a rate of 700 gallons per hour.

Q: How can we model this situation mathematically?

A: We can use the concept of rate of change to model this situation. The rate at which the pool is being drained is 700 gallons per hour, which means that the number of gallons remaining in the pool decreases by 700 gallons every hour. We can represent this relationship using the equation:

dgdt=−700\frac{dg}{dt} = -700

where $\frac{dg}{dt}$ represents the rate of change of the number of gallons remaining in the pool with respect to time.

Q: How do we solve the differential equation?

A: To solve the differential equation, we need to integrate both sides with respect to time:

∫dgdtdt=∫−700dt\int \frac{dg}{dt} dt = \int -700 dt

g(t)=−700t+Cg(t) = -700t + C

where $C$ is the constant of integration.

Q: What is the initial condition?

A: We know that initially, the pool contains 28,000 gallons of water. We can use this information to find the value of the constant $C$:

g(0)=28,000g(0) = 28,000

Substituting $t = 0$ into the equation, we get:

g(0)=−700(0)+Cg(0) = -700(0) + C

C=28,000C = 28,000

Q: What is the equation that models the situation?

A: The equation that models the situation is:

g(t)=−700t+28,000g(t) = -700t + 28,000

This equation represents the relationship between the number of gallons remaining in the pool and the amount of time the pump has been operating.

Q: How can we use this equation to find the number of gallons remaining in the pool at any given time?

A: We can use this equation to find the number of gallons remaining in the pool at any given time by substituting the value of $t$ into the equation. For example, if the pump has been operating for 4 hours, we can substitute $t = 4$ into the equation:

g(4)=−700(4)+28,000g(4) = -700(4) + 28,000

g(4)=−2,800+28,000g(4) = -2,800 + 28,000

g(4)=25,200g(4) = 25,200

This means that after 4 hours, there will be 25,200 gallons of water remaining in the pool.

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

A: This scenario has many real-world applications, such as:

  • Calculating the time it takes to drain a swimming pool
  • Determining the amount of water remaining in a pool after a certain amount of time
  • Modeling the behavior of fluids in various engineering applications

In this article, we answered some of the most frequently asked questions related to the scenario of a 28,000-gallon swimming pool being drained using a pump that empties 700 gallons per hour. We provided additional insights and clarifications, and highlighted the importance of mathematical modeling in real-world applications.

  • [1] Calculus: Early Transcendentals, 8th edition, by James Stewart
  • [2] Differential Equations and Dynamical Systems, 4th edition, by Lawrence Perko

For more information on mathematical modeling and differential equations, we recommend the following resources:

  • [1] Khan Academy: Differential Equations
  • [2] MIT OpenCourseWare: Differential Equations
  • [3] Wolfram MathWorld: Differential Equations