Problem 35 - Calculus Single Variable

Question

A filled reservoir is in the shape of a frustum of an inverted cone as shown in Figure \(9 .\) The radius of the circular base of the cone is 100 feet. The radius of the circular top of the frustum and the depth at the center are both 50 feet. A pump that floats on the surface of the water pumps water to the top of the reservoir, at which point the water runs off. How much work is done emptying the reservoir?

Step-by-Step Solution

Answer

The total work done emptying the reservoir is approximately \( 675\times 10^9\pi \) foot-pounds.

Step 1: Determine Frustum Volume Formula

To solve for the work done in emptying the reservoir, we start by understanding its shape. A frustum of an inverted cone has a volume calculated using the formula: \[ V = \frac{1}{3} \pi h \left(R_1^2 + R_1 R_2 + R_2^2\right) \] where \( h \) is the height, \( R_1 \) is the radius of the bottom base, and \( R_2 \) is the radius of the top base.

Step 2: Calculate Frustum Dimensions

Given that the radius of the bottom base \( R_1 = 100 \text{ ft} \) and the radius of the top base \( R_2 = 50 \text{ ft} \), the height \( h \) is also given as 50 feet. Substitute these values into the frustum volume formula to find the volume.

Step 3: Compute the Frustum Volume

Substituting the given dimensions into the formula, we have: \[ V = \frac{1}{3} \pi (50) \left(100^2 + 100 \cdot 50 + 50^2\right) \]. Calculating inside the parentheses: \( 100^2 = 10000 \), \( 100 \cdot 50 = 5000 \), and \( 50^2 = 2500 \), so the total is 17500. Hence, \[ V = \frac{1}{3} \pi \cdot 50 \cdot 17500 \]. This simplifies to \[ V = 875000\pi \text{ cubic feet} \].

Step 4: Calculate Work Done

Work is the integration of the force applied over the distance. The differential element of work \( dW \) in lifting \( dV \) is expressed as: \[ dW = \rho g y dV \] with \( \rho \) as the density of water, \( g = 9.8 \text{ m/s}^2 \) as the acceleration due to gravity, and \( y \) is the distance each water slice has to be lifted. \( \rho \) for water in English units is \( 62.4 \text{ lb/ft}^3 \).

Step 5: Set Up the Integration for Work

The range for \( y \) runs from 0 to 50. Therefore, the work done \( W \) is given by \( W = \int_0^{50} 62.4 \times 9.8 \times y \times 875000 \pi dy \), adjusting the total volume due to distance \( y \) each slice is moved.

Step 6: Solve the Integration

Evaluate the integral \[ W = 62.4 \times 9.8 \times 875000 \pi \int_0^{50} y \, dy \]. Integrating \( y \) from 0 to 50 results in: \[ \int_0^{50} y \, dy = \left[ \frac{y^2}{2} \right]_0^{50} = \frac{50^2}{2} - \frac{0^2}{2} = 1250 \].

Step 7: Final Algebraic Calculation of Work

After solving the integral, plug back into the expression for work: \[ W = 62.4 \times 9.8 \times 875000 \pi \times 1250 \]. Calculating this expression gives the total work done in foot-pounds.

Key Concepts

Frustum Volume

A frustum is a section of a solid, created when a top portion is cut off, usually parallel to the base. Calculating the volume of a frustum of an inverted cone involves knowing both the top and bottom radii of the cone segments. The formula is given by:\[ V = \frac{1}{3} \pi h \left(R_1^2 + R_1 R_2 + R_2^2\right) \]where:

• \(R_1\) is the radius of the bottom base.
• \(R_2\) is the radius of the top base.
• \(h\) is the height of the frustum.

To find the volume accurately, substitute the given dimensions into this formula. For this specific exercise, with bottom radius 100 feet, top radius 50 feet, and height 50 feet, the volume is resolved to \(875000\pi\) cubic feet.

Integration

In calculus, integration is used to calculate quantities such as areas, volumes, and in this case, work. Here, integration is used to sum up the infinitesimal slices of the frustum as they are lifted out of the reservoir. The differential work done, expressed as \(dW = \rho g y dV\), involves:

• \(\rho\), the density of the fluid (water in this case), which is 62.4 lb/ft³ in English units.
• \(g\), the acceleration due to gravity which is approximately 9.8 m/s².
• \(y\), the vertical position of each slice.

The integral is set up over the height of the frustum: \(\int_0^{50} \rho g y dV\). Calculating this integral incorporates all the slices that make up the frustum, concluding with the total work required to move water from the bottom to the top.

Physics Applications

Physics concepts such as work and energy allow us to determine physical quantities based on interactions. In this scenario, calculating the work to pump water is essential for engineering and real-world applications.

• **Work** is the product of force applied over a distance: \[ W = \int F(y) \cdot dy \]
• **Force** in this context is due to gravity, which is influenced by the mass of the water and gravitational acceleration.

Using these principles, engineers can predict how much energy (or work) is necessary to perform tasks such as emptying reservoirs, ensuring efficient design and operation. This specific task was derived from work = force x distance over the height of the water in the frustum.

Inverted Cone Geometry

The shape of the inverted cone and its frustum plays a pivotal role in solving the given problem. A complete inverted cone would normally have a single vertex downward and a wide circular opening at the top. However, a frustum excludes the vertex and maintains parallel circular bases at different radii.

• The bottom base's larger radius is denoted as \(R_1\), while the smaller top base radius is \(R_2\).
• The height of the frustum provides the perpendicular distance between the two circular faces.

Understanding this geometry is crucial, as it affects how we apply the volume formula and calculate the work necessary to account for lifting all slices of water. The inversion noted here primarily affects how gravity acts, which is consistent for analyzing a reservoir or similar applications.