A Tub Filled With 50 Quarts Of Water Empties At A Rate Of 2.5 Quarts Per Minute. Let $w$ Be The Quarts Of Water Left In The Tub, And $t$ Be The Time In Minutes.$\[ \begin{tabular}{|c|c|} \hline \text{Time } (t) & \text{Quarts Of

Introduction

In this article, we will explore the concept of rate of change and how it can be used to model real-world situations. We will consider a tub filled with 50 quarts of water that empties at a rate of 2.5 quarts per minute. Our goal is to determine the amount of water left in the tub at any given time.

Mathematical Modeling

Let $w$ be the quarts of water left in the tub, and $t$ be the time in minutes. We can model the situation using the following differential equation:

$$\frac{dw}{dt} = -2.5$$

where $\frac{dw}{dt}$ represents the rate of change of the amount of water in the tub with respect to time.

Solving the Differential Equation

To solve the differential equation, we can use the following method:

$$\int \frac{dw}{dt} dt = \int -2.5 dt$$

$$w(t) = -2.5t + C$$

where $C$ is the constant of integration.

Initial Condition

We are given that the tub initially contains 50 quarts of water. This can be represented by the following initial condition:

$$w(0) = 50$$

Solving for the Constant of Integration

We can use the initial condition to solve for the constant of integration $C$. Substituting $t = 0$ and $w(0) = 50$ into the equation, we get:

$$50 = -2.5(0) + C$$

$$C = 50$$

Final Solution

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

$$w(t) = -2.5t + 50$$

This is the final solution to the differential equation.

Graphical Representation

The graph of the solution is a straight line with a negative slope. The y-intercept is 50, which represents the initial amount of water in the tub. The x-intercept is 20, which represents the time it takes for the tub to empty.

Discussion

The differential equation $\frac{dw}{dt} = -2.5$ represents the rate of change of the amount of water in the tub with respect to time. The solution $w(t) = -2.5t + 50$ represents the amount of water left in the tub at any given time.

Conclusion

In this article, we have explored the concept of rate of change and how it can be used to model real-world situations. We have considered a tub filled with 50 quarts of water that empties at a rate of 2.5 quarts per minute. Our goal was to determine the amount of water left in the tub at any given time. We have used a differential equation to model the situation and have solved for the amount of water left in the tub at any given time.

Table of Values

Example Problems

1. If the tub empties at a rate of 2.5 quarts per minute, how much water will be left in the tub after 10 minutes?
2. If the tub initially contains 75 quarts of water, how much water will be left in the tub after 15 minutes?

Answer Key

1. 30
2. 30

References

• [1] "Differential Equations and Their Applications" by Martin Braun
• [2] "Calculus" by Michael Spivak

Conclusion

In conclusion, the differential equation $\frac{dw}{dt} = -2.5$ represents the rate of change of the amount of water in the tub with respect to time. The solution $w(t) = -2.5t + 50$ represents the amount of water left in the tub at any given time. We have used a differential equation to model the situation and have solved for the amount of water left in the tub at any given time.