Problem 22 - Single Variable Calculus

Question

A tank having the shape of a right circular cylinder with a radius of \(5 \mathrm{ft}\) and a height of \(6 \mathrm{ft}\) is filled with water weighing \(62.4 \mathrm{lb} / \mathrm{ft}^{3}\). Find the work required to empty the tank by pumping the water out of the tank through a pipe that extends to a height of \(2 \mathrm{ft}\) beyond the top of the tank.

Step-by-Step Solution

Answer

The work required to empty the tank by pumping the water out through the pipe is approximately \(15,624\pi \text{ ft-lb}\).

Step 1: Determine the distance the water must be pumped vertically at each level of the tank

For each level in the tank, we need to calculate the vertical distance the water has to travel. At the bottom of the tank, the water must travel the full height of 6 ft as well as the additional 2 ft beyond the top, for a total of 8 ft. As we move up the tank, this distance decreases, and at the top of the tank, water only needs to travel the additional 2 ft.

Step 2: Divide the tank into thin horizontal layers

We will divide the tank into thin horizontal layers to calculate the work required to lift the water at each level. Let's consider a layer of water with height \(\Delta h\) at a distance \(h\) from the bottom of the tank. The volume of this layer is \(\Delta V = \pi (5^2) \Delta h\).

Step 3: Calculate the work done for each layer

The weight of the water is given as \(62.4 \text{ lb/ft}^3\). Therefore, the weight of the water in the layer is \(\Delta W = 62.4 \Delta V = 62.4 \pi (5^2) \Delta h\). To pump the water at this level through the pipe, we need to lift it \((6- h + 2)\) ft. The work required for this lifting is given by \(\Delta U = \Delta W(6 - h + 2) = 62.4\pi (5^2) \Delta h (8 - h)\).

Step 4: Integrate the work done for all layers

To find the total work required to empty the tank, we will integrate the work done for all the layers from the bottom to the top of the tank. That is: $$ U = \int_{0}^{6} 62.4\pi (5^2)(8 - h)dh $$

Step 5: Calculate the integral

Now, we will evaluate the integral: \begin{align*} U &= 62.4\pi (5^2) \int_{0}^{6} (8 - h)dh \\ &= 62.4\pi (5^2) \left[8h - \frac{1}{2}h^2\right]_{0}^{6} \\ &= 62.4\pi (5^2) \left([8(6) - \frac{1}{2}(6)^2] - [0]\right) \\ &= 62.4\pi (5^2) (48 - 18) \\ &= 15,624\pi \text{ ft-lb} \end{align*} So, the work required to empty the tank by pumping the water out through the pipe is approximately \(15,624\pi \text{ ft-lb}\).

Key Concepts

Work done in physics

Imagine you have a full tank of water, and your task is to empty it. To accomplish this, you need to apply a force to move the water upwards and out of the tank. The concept here revolves around 'work done' in physics, which means the amount of energy required to perform this task.

Work, in a physical context, is calculated by multiplying the force exerted by the distance over which this force acts. In mathematical terms, it's expressed as:

\[ W = F imes d \], where:

• \(W\) is the work done,
• \(F\) is the force applied, and
• \(d\) is the distance the object is moved.

In our exercise with the tank, the water is pumped up over a certain height, and thus, the force of gravity acting against this movement helps us decide how much work is needed. The result is a quantifiable measure of energy used in pumping water from different layers of the tank to the top.

Definite integrals

Definite integrals serve as a tool for calculating the total sum of an infinite number of infinitesimally small quantities. This might sound complex, but it's a way to add up little bits to find a comprehensive outcome.

In calculus, a definite integral calculates the net area between the curve of a function and the axis over an interval \([a, b]\). The formula is represented as:

\[ \int_{a}^{b} f(x) \, dx \]In our tank example, each infinitesimal segment of water at height \(h\) is considered and the work required to lift that segment is calculated. By taking the integral over the height of the tank (from 0 to 6 ft), we find the total work required to empty the tank.

The transformation of infinitesimal calculations into a comprehensive sum showcases the powerful application of definite integrals in determining real-world quantities, like the work needed to pump the water out of a cylindrical tank.

Volume of revolution

The concept of volume of revolution is about finding the volume of a three-dimensional object created by rotating a two-dimensional shape around an axis. Although this is not directly applied in our exercise, understanding this principle is helpful when dealing with cylindrical tanks.

A right circular cylinder, the shape of our tank, can be imagined as being generated by rotating a rectangle around one of its sides (the height), forming the circular cross-section. Using calculus, the volume of this solid can be determined by integrating across this circular cross-section:

\[ V = \pi r^2 h \], where:

• \(V\) is the volume,
• \(r\) is the radius, and
• \(h\) is the height of the cylinder.

Understanding this concept aids in visualizing and setting up problems where the shape of the body is crucial, especially relevant in fluid dynamics and when calculating the hydrostatic force and pressure in different sections of the tank.

Physics applications of calculus

Calculus proves its worth in many physical applications, facilitating the analysis and solution of problems that include changing quantities and rates of change.

In the tank exercise, calculus helps compute the work done by integrating the forces over distances. This seamless interweaving of calculus and physics aids in analyzing complex systems—where variable forces require sophisticated treatment.

Some noteworthy applications include:

• Calculating work done, as we saw in this problem, where water is pumped out of a tank with varying effort based on water height.
• Modeling the motion of objects under non-uniform forces like gravitational fields.
• Determining the rate of its volume change when objects rotate or flow.

These processes allow scientists and engineers to predict system behavior with great accuracy, showcasing the indispensable utility of calculus in the physical world.