Problem 30 - Calculus Single Variable

Question

A swimming pool has rectangular base with side lengths of 16 and 30 feet. The pool is 10 feet deep but not completely filled: the depth of the water it contains is 8 feet. A pump floats on the surface of the water and pumps the water to the top of the pool, at which point the water runs off. How much work does the pump perform in emptying the pool?

Step-by-Step Solution

Answer

The pump performs 478,080 ft-lb of work to empty the pool.

Step 1: Calculate the Volume of Water

The pool is a rectangular prism with dimensions 16 ft by 30 ft, and it contains water to a depth of 8 ft. To find the volume of water, multiply the base area by the water depth: \( V = 16 \times 30 \times 8 \).

Step 2: Identify the Weight of the Water

Water has a weight density of 62.4 pounds per cubic foot. Multiply the water volume by this density to determine the total weight: \( W = V \times 62.4 \).

Step 3: Calculate the Average Height to Lift Water

Since the pump lifts water from the surface at an 8-foot depth to the top of the pool, each infinitesimal slice of water elevated from depth \( y \) must be accounted for. The average height from the water surface to the top of the pool is 2 feet (average from 0 to 2).

Step 4: Calculate the Work Done

The work done is the integral of weight lifted times the lifting height over the depth interval from 8 to 10 feet. Calculate \( W = \int_{8}^{10} (62.4 \cdot 16 \cdot 30) (10-y) \, dy \).

Step 5: Evaluate the Integral

To solve the integral \( W = 62.4 \times 480 \int_{8}^{10} (10-y) \, dy \), find the antiderivative and calculate its values at the limits 8 and 10: \( W = 62.4 \times 480 \times [-\frac{y^2}{2} + 10y]_{8}^{10} \).

Step 6: Find the Final Result

Compute the result using the evaluated antiderivative: \( W = 62.4 \times 480 \times [100 - 64 - 80 + 40] = 62.4 \times 480 \times 16 \).

Step 7: Compute the Product

Calculate \( 62.4 \times 480 \times 16 \) to find the work done: \( W = 478080 \text{ ft-lb} \).

Key Concepts

Volume Calculation

To solve the problem of how much work the pump needs to perform, we begin by understanding the concept of volume calculation. Imagine the swimming pool as a giant rectangular box. The base of the pool measures 16 feet by 30 feet. To calculate the volume of water, think about filling this box up to a specific height. In this case, the height is 8 feet, as the pool is not entirely filled.
It’s helpful to break down the volume formula:

• Base area = Length × Width = 16 ft × 30 ft
• Volume = Base area × Height of water = 16 ft × 30 ft × 8 ft

This calculation tells us how much space the water takes up in the pool.

Integration in Physics

When dealing with physics problems involving lifting or moving substances, integration becomes a crucial concept. In this exercise, integration helps calculate the total work the pump does by adding up tiny amounts of work needed to lift each slice of water.
The height from which each slice of water is lifted varies, requiring an integral. To realize the task:

• Consider the infinitesimal slices of water at different heights in the pool's depth range of 8 feet to 10 feet.
• Integrate over this range: \[ \text{Work} = \int_{8}^{10} (62.4 \cdot 16 \cdot 30) (10-y) \, dy \]

This integral sums up work done by lifting all the water from the bottom to the top of the pool.

Weight Density

Weight density is a key term when working with substances like water where we need to relate volume to weight. In this case, it is the weight of a unit volume of water, expressed in pounds per cubic foot.
Water weight density is typically about 62.4 pounds per cubic foot. This means, for every cubic foot of water, it weighs 62.4 pounds.

• To find the total weight of water, multiply its volume by the weight density: \[ W = V \times 62.4 \]

Understanding weight density helps us move from just knowing the volume of water to knowing how heavy it is, crucial for calculating work.

Antiderivative Evaluation

To calculate the work involved in lifting the water, evaluating the antiderivative of the integral expression is necessary. This step essentially involves finding a function whose derivative gives us the integrand.

• Start with the integral expression: \[ \int_{8}^{10} (62.4 \cdot 480) (10-y) \, dy \]Breaking it down, the antiderivative for \((10-y)\) is \(-\frac{y^2}{2} + 10y\).
• Upon substituting the limits of integration (8 to 10), calculate:\[ W = 62.4 \times 480 \times \left[ -\frac{y^2}{2} + 10y \right]_{8}^{10} \]

Through this evaluation, we gather the total amount of work required for the water to move upwards, arriving at the final numerical solution through further calculations.