Problem 250 - Calculus Volume 1

Question

For the following exercises, find the work done. [T] How much work is required to pump out a swimming pool if the area of the base is 800 \(\mathrm{ft}^{2}\) , the water is 4 \(\mathrm{ft}\) deep, and the top is 1 \(\mathrm{ft}\) above the water level? Assume that the density of water is 62 \(\mathrm{lb} / \mathrm{ft}^{3}\) .

Step-by-Step Solution

Answer

The work required is 595200 foot-pounds.

Step 1: Understanding the Problem

We need to find the work done in pumping water out of a swimming pool. We'll consider the physical quantities given: the base area of the pool is 800 ft², the depth of the water is 4 ft, and the top of the pool is 1 ft above the water level. The density of water is given as 62 lb/ft³. The work done is calculated using the concept of force times distance.

Step 2: Setting Up the Force Calculation

The force required to move a given volume of water depends on the water's weight. First, find the volume of a slice of water with a small thickness \(\Delta y\) at a depth \y\ using base area times thickness: \V = 800 \, \Delta y\, \, \mathrm{ft}^{3}\. The weight \w\ of this slice is density times volume: \w = 62 \, \mathrm{lb/ft}^3 \, \times 800 \, \Delta y = 49600 \, \Delta y\, \, \mathrm{lb}\.

Step 3: Establishing the Work Formula

The work done to move a slice from depth \y\ up to 1 ft out of the water involves moving it a distance \(5 - y\) ft (since it rises from depth \(y\) to height 5 ft where pumping out occurs). Work \(ΔW\) for this slice is weight times the distance moved: \(ΔW = 49600 \, Δy (5 - y)\) foot-pounds.

Step 4: Calculating Total Work Done

To find the total work done, integrate over the depth of the water from 0 to 4 ft:\[ W = \int_{0}^{4} 49600 (5 - y) \, dy. \]

Step 5: Solving the Integral

Solve the integral:\[W = 49600 \int_{0}^{4} (5 - y) \, dy = 49600 \left[5y-\frac{y^2}{2}\right]_{0}^{4}.\]Evaluate:\[W = 49600 \left[(5(4) - \frac{16}{2}) - (5(0) - \frac{0^2}{2})\right]= 49600 (20 - 8) = 49600 \times 12 = 595200 \, \text{foot-pounds}.\]

Step 6: Concluding the Solution

The work required to pump all the water out of the swimming pool is 595200 foot-pounds. This takes into account the force needed to lift the water to the given height.

Key Concepts

Calculating Work Done

In physics, work is defined as the amount of energy transferred by a force over a distance. When discussing work done, it's crucial to understand the relationship between force, distance, and direction. Here, we're looking at an application where we need to pump water out of a swimming pool. We calculate work done using the formula:

• Work (W) = Force (F) x Distance (D)

In this scenario, the force is equivalent to the weight of the water that needs to be moved. Weight is a product of the volume of water and the density of water. The distance is how far each infinitesimally thin slice of water must be lifted.
For the swimming pool example, the total work required is calculated by segmenting the volume of water into horizontal slices, calculating the weight per slice, and then determining the distance each slice must be moved. Since each slice travels a different distance, the work for each one varies. It’s these small segments summed up, often using integration, that give us the total work done.

Integration in Calculus

Integration in calculus allows us to compute a vast array of values. For our swimming pool example, integration was used to add up the work done on all the tiny slices of water. This process nicely sums the contributions of each slice to calculate the total work required.
Integration deals with the concept of accumulative addition, which is perfect for finding totals from a continuously varying quantity. In this context, integration takes:

• A function that represents work done on a slice of water, such as \(49600(5-y)\)
• The bounds from where the slices contribute (0 ft to 4 ft)

Finally, we evaluate the integral to determine the total work. This provides a real-world application of calculus beyond theoretical exercises, showcasing its usefulness in problem-solving.

Density and Weight Calculations

The concept of density is core to calculating weight, especially involving fluids like water. Density is the mass per unit volume of a substance, and in our exercise, the density of water is given as 62 lb/ft³. This means that for each cubic foot of water, it weighs 62 pounds.
When calculating the work done, knowing the density helps because:

• You can find the weight of a specific volume of water.
• This weight then directly allows you to calculate the force needed to move it.

The solution takes a specific slice of the pool's water, estimating its volume, and multiplies it by the density to find the slice's weight. This way, density helps bridge the gap between abstract volume measurements and practical weight considerations necessary for computing work.

Applications of Integrals

Integrals have a broad range of uses in various fields, from physics to finance. In the context of our exercise, this involved using integrals to compute the amount of work required to pump water from a pool.
Applications of integrals like this extend beyond to:

• Calculating areas under curves, which is fundamental in statistics and probability.
• Determining volumes and surface areas of more complex 3D shapes.
• Solving differential equations, key in modeling real-world systems.

Through this specific application, students can appreciate integrals as powerful tools for solving dynamic and spatial problems, showing how mathematical concepts translate into practical solutions in physics and engineering.