Problem 24 - Calculus Early Transcendentals: Pearson New International Edition

Question

The interior of an open cylindrical tank is 12 feet in diameter and 8 feet deep. The bottom is copper and the sides are steel. Use differentials to find approximately how many gallons of waterproofing paint are needed to apply a \(0.05\) -inch coat to the steel part of the inside of the tank \((1\) gallon \(\approx 231\) cubic inches \() .\)

Step-by-Step Solution

Answer

Approximately 9.4 gallons of waterproofing paint are needed.

Step 1: Find the surface area of the steel part

The steel part is the sides of the cylinder. The formula for the lateral surface area of a cylinder is \( A = 2\pi rh \), where \( r \) is the radius and \( h \) is the height. Here, \( r = \frac{12}{2} = 6 \) feet and \( h = 8 \) feet. Converting feet to inches (since the coat thickness is given in inches), we have \( r = 72 \) inches and \( h = 96 \) inches. Therefore, \( A = 2\pi \times 72 \times 96 = 13824\pi \) square inches.

Step 2: Determine the volume of the paint layer

The volume of paint needed for a 0.05-inch coat is approximately equal to the surface area times the thickness of the coat. Using differential approximation, the volume \( V \approx A \times \Delta h = 13824\pi \times 0.05 \), where \( A = 13824\pi \) square inches, and \( \Delta h = 0.05 \) inches. Therefore, \( V \approx 691.2\pi \approx 2171.01 \) cubic inches.

Step 3: Convert cubic inches to gallons

Since 1 gallon is approximately 231 cubic inches, the volume in gallons is \( V_g = \frac{2171.01}{231} \approx 9.4 \) gallons.

Key Concepts

Lateral Surface Area

The lateral surface area of a cylinder refers to the area of its side, excluding the top and bottom. This area is calculated by unrolling the cylindrical surface into a rectangle. Imagine slicing the cylinder vertically and laying it flat to create a rectangle where:

• The length is equal to the circumference of the circle (the base of the cylinder), which can be calculated as \(2\pi r\).
• The width is the height \(h\) of the cylinder.

Thus, the formula to find the lateral surface area \(A\) is \(A = 2\pi rh\). In this exercise, about the cylindrical tank, the radius is given as 6 feet and the height as 8 feet. It's important to ensure all measurements are in the same units. Hence, we convert feet to inches because the coat thickness is provided in inches. This leads to the calculation: \(A = 2\pi \times 72 \times 96 = 13824\pi\) square inches. Understanding this concept helps in effectively calculating the area of cylindrical surfaces when dealing with any such problems.

Volume Approximation

Volume approximation via differentials is a useful technique in calculus for estimating small changes in volume due to adjustments in dimensions. Here, the approximation links the lateral surface area of the cylinder and the thickness of the coat of paint needed. If we think about adding an extremely thin layer of material over a large area, imagining that the change in volume is essentially the product of the surface area and the thickness is logical.To calculate the amount of waterproofing paint, we approximate the volume \(V\) as the product of the surface area \(A\) and the layer thickness \(\Delta h\), that is, \(V \approx A \times \Delta h\). In the exercise, we calculated the area to be \(13824\pi\) square inches and the thickness \(\Delta h\) to be 0.05 inches. Using these values, the volume of paint required is approximated as \(V \approx 13824\pi \times 0.05 \approx 2171.01\) cubic inches. This method showcases the power of differential calculus in making practical, real-world estimates.

Unit Conversion

Unit conversion is crucial when working across systems of measurement or dimensions. In this problem, one starts with dimensions in feet for the radius and height of the cylinder.However, as the coat of paint thickness is given in inches, it's necessary to convert all measurements to inches to maintain consistency in calculations.Conversion is straightforward: one foot equals twelve inches. Thus, for the radius:

• 6 feet translates to \(6 \times 12 = 72\) inches.

Similarly, for the height:

• 8 feet becomes \(8 \times 12 = 96\) inches.

Working with the same units is mandatory for accurate calculations, as it ensures that formulas operate correctly without unexpected distortions arising from inconsistent units. Similarly, converting the final volume from cubic inches to gallons is needed, since gallons are a more practical unit for measuring liquids like paint. Here, remembering that 1 gallon equals approximately 231 cubic inches is useful.

Cylinder Geometry

Understanding cylinder geometry is foundational for working effectively with cylindrical shapes in mathematics and engineering. A cylinder consists of two parallel circular bases and a curved side. The geometry is defined primarily by its radius \(r\) and height \(h\).In the context of this exercise, the cylindrical tank is open, implying that one of the circular bases is absent—the copper base in this case remains, while the steel sides require coating. Calculating dimensions like radius from diameter is essential: the radius equals half the diameter.Understanding these parameters aids in calculating other properties, such as:

• The lateral surface area, which requires knowledge of base circumference (\(2\pi r\)) and height \(h\).
• Knowing how to visualize or imagine the surfaces involved, distinguishing between the different parts such as the base and the side, helps see which parts need treatment like painting.

Mastering cylinder geometry offers insight into both simple calculations and complex engineering tasks involving volume and surface areas.