Raj's Bathtub Is Clogged And Is Draining At A Rate Of 1.5 Gallons Of Water Per Minute. The Table Shows That The Amount Of Water Remaining In The Bathtub, Y Y Y , Is A Function Of The Time In Minutes, X X X , That It Has Been

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Raj's Bathtub Clog: A Mathematical Exploration of Water Drainage

Raj's bathtub is clogged, and water is draining at a rate of 1.5 gallons per minute. This situation presents an opportunity to explore the mathematical relationship between the amount of water remaining in the bathtub and the time it has been draining. In this article, we will delve into the world of mathematics to understand the underlying principles governing this phenomenon.

The table below shows the amount of water remaining in the bathtub, yy, as a function of the time in minutes, xx, that it has been draining.

Time (minutes) Water Remaining (gallons)
0 120
5 90
10 60
15 30
20 0

To understand the relationship between the amount of water remaining in the bathtub and the time it has been draining, we can use a mathematical model. Let's assume that the rate at which water is draining is constant, which is 1.5 gallons per minute. We can use the concept of exponential decay to model the situation.

Exponential Decay

Exponential decay is a process where the amount of a substance decreases over time at a rate proportional to its current amount. In this case, the amount of water remaining in the bathtub decreases at a rate proportional to its current amount. We can represent this using the equation:

y=y0eβˆ’kty = y_0 e^{-kt}

where:

  • yy is the amount of water remaining in the bathtub at time tt
  • y0y_0 is the initial amount of water in the bathtub (120 gallons)
  • kk is the decay rate (which we need to determine)
  • tt is the time in minutes

Determining the Decay Rate

To determine the decay rate kk, we can use the data from the table. We can choose two points from the table and use them to calculate the decay rate. Let's choose the points (5, 90) and (10, 60).

Using the equation y=y0eβˆ’kty = y_0 e^{-kt}, we can write:

90=120eβˆ’k(5)90 = 120 e^{-k(5)}

60=120eβˆ’k(10)60 = 120 e^{-k(10)}

We can solve these equations simultaneously to determine the value of kk.

Solving for k

To solve for kk, we can divide the two equations:

9060=120eβˆ’k(5)120eβˆ’k(10)\frac{90}{60} = \frac{120 e^{-k(5)}}{120 e^{-k(10)}}

Simplifying the equation, we get:

32=e5k\frac{3}{2} = e^{5k}

Taking the natural logarithm of both sides, we get:

ln⁑(32)=5k\ln\left(\frac{3}{2}\right) = 5k

Solving for kk, we get:

k=ln⁑(32)5k = \frac{\ln\left(\frac{3}{2}\right)}{5}

Calculating the Decay Rate

Now that we have determined the value of kk, we can calculate the decay rate.

k=ln⁑(32)5β‰ˆ0.139k = \frac{\ln\left(\frac{3}{2}\right)}{5} \approx 0.139

The Mathematical Model

Now that we have determined the decay rate, we can write the mathematical model for the situation.

y=120eβˆ’0.139ty = 120 e^{-0.139t}

The mathematical model represents the amount of water remaining in the bathtub as a function of the time it has been draining. The model shows that the amount of water remaining in the bathtub decreases exponentially over time.

In this article, we explored the mathematical relationship between the amount of water remaining in Raj's bathtub and the time it has been draining. We used the concept of exponential decay to model the situation and determined the decay rate. The mathematical model represents the amount of water remaining in the bathtub as a function of the time it has been draining.

In the future, we can use this mathematical model to predict the amount of water remaining in the bathtub at any given time. We can also use this model to explore other scenarios, such as a bathtub with a different initial amount of water or a bathtub with a different drainage rate.

The following is a list of equations used in this article:

  • y=y0eβˆ’kty = y_0 e^{-kt}
  • 90=120eβˆ’k(5)90 = 120 e^{-k(5)}
  • 60=120eβˆ’k(10)60 = 120 e^{-k(10)}
  • 9060=120eβˆ’k(5)120eβˆ’k(10)\frac{90}{60} = \frac{120 e^{-k(5)}}{120 e^{-k(10)}}
  • ln⁑(32)=5k\ln\left(\frac{3}{2}\right) = 5k
  • k=ln⁑(32)5k = \frac{\ln\left(\frac{3}{2}\right)}{5}
  • y=120eβˆ’0.139ty = 120 e^{-0.139t}
    Raj's Bathtub Clog: A Mathematical Exploration of Water Drainage - Q&A

In our previous article, we explored the mathematical relationship between the amount of water remaining in Raj's bathtub and the time it has been draining. We used the concept of exponential decay to model the situation and determined the decay rate. In this article, we will answer some frequently asked questions related to this topic.

Q: What is exponential decay?

A: Exponential decay is a process where the amount of a substance decreases over time at a rate proportional to its current amount. In the context of Raj's bathtub, the amount of water remaining in the bathtub decreases at a rate proportional to its current amount.

Q: How do you determine the decay rate?

A: To determine the decay rate, we can use the data from the table. We can choose two points from the table and use them to calculate the decay rate. Let's choose the points (5, 90) and (10, 60). We can use the equation y=y0eβˆ’kty = y_0 e^{-kt} to determine the decay rate.

Q: What is the mathematical model for the situation?

A: The mathematical model for the situation is y=120eβˆ’0.139ty = 120 e^{-0.139t}. This equation represents the amount of water remaining in the bathtub as a function of the time it has been draining.

Q: Can you explain the concept of exponential decay in simpler terms?

A: Exponential decay is like a snowball rolling down a hill. The snowball starts big and rolls down the hill, getting smaller and smaller as it goes. The rate at which the snowball gets smaller is proportional to its current size. In the context of Raj's bathtub, the amount of water remaining in the bathtub decreases at a rate proportional to its current amount.

Q: How can you use this mathematical model in real-life situations?

A: This mathematical model can be used to predict the amount of water remaining in a bathtub at any given time. It can also be used to explore other scenarios, such as a bathtub with a different initial amount of water or a bathtub with a different drainage rate.

Q: What are some limitations of this mathematical model?

A: One limitation of this mathematical model is that it assumes a constant drainage rate. In reality, the drainage rate may vary depending on factors such as the size of the bathtub, the type of drain, and the amount of water in the bathtub.

Q: Can you provide more examples of exponential decay in real-life situations?

A: Exponential decay can be seen in many real-life situations, such as:

  • Radioactive decay: The amount of radioactive material decreases over time at a rate proportional to its current amount.
  • Population growth: The population of a species may decrease over time at a rate proportional to its current size.
  • Chemical reactions: The amount of a substance may decrease over time at a rate proportional to its current amount.

In this article, we answered some frequently asked questions related to the mathematical relationship between the amount of water remaining in Raj's bathtub and the time it has been draining. We hope this article has provided a better understanding of the concept of exponential decay and its applications in real-life situations.

The following is a list of equations used in this article:

  • y=y0eβˆ’kty = y_0 e^{-kt}
  • 90=120eβˆ’k(5)90 = 120 e^{-k(5)}
  • 60=120eβˆ’k(10)60 = 120 e^{-k(10)}
  • 9060=120eβˆ’k(5)120eβˆ’k(10)\frac{90}{60} = \frac{120 e^{-k(5)}}{120 e^{-k(10)}}
  • ln⁑(32)=5k\ln\left(\frac{3}{2}\right) = 5k
  • k=ln⁑(32)5k = \frac{\ln\left(\frac{3}{2}\right)}{5}
  • y=120eβˆ’0.139ty = 120 e^{-0.139t}