Problem 62 - Algebra and Trigonometry

Question

Rain Gutter A rain gutter is to be constructed from a metal sheet of width 30 \(\mathrm{cm}\) by bending up one-third of the sheet on each side through an angle \(\theta\) . (a) Show that the cross-sectional area of the gutter is modeled by the function $$A(\theta)=100 \sin \theta+100 \sin \theta \cos \theta$$ (b) Graph the function \(A\) for \(0 \leq \theta \leq \pi / 2\) (c) For what angle \(\theta\) is the largest cross-sectional area achieved?

Step-by-Step Solution

Answer

The maximum area occurs at \( \theta = \frac{\pi}{4} \). The area function describes the gutter's cross-section efficiently.

Step 1: Understand the Problem

We start by considering the metal sheet's width of 30 cm, which is divided into three equal parts. The central part remains horizontal, while the side parts are bent upward at an angle \( \theta \). Each side part is 10 cm in width.

Step 2: Establish the Geometry

Visualize the gutter as a shape formed by bending up the sides of a rectangle. The base is 10 cm wide, and the sides each become the hypotenuse of a right triangle with a height and base of equal length, \(10 \sin \theta\) and \(10 \cos \theta\), respectively.

Step 3: Derive the Area Formula

The cross-sectional area, \( A(\theta) \), is a combination of the vertical height and the effective base altered by \( \cos \theta \). Thus, \( A(\theta) = b \times 10 \sin \theta = (10 \cos \theta + 10) \times 10 \sin \theta \). Simplifying leads to \( A(\theta) = 100 \sin \theta + 100 \sin \theta \cos \theta \).

Step 4: Plot the Function

To graph \( A(\theta) = 100 \sin \theta + 100 \sin \theta \cos \theta \) over the interval \([0, \pi/2]\), note that \( A(\theta) \) describes how the area changes with the angle \( \theta \). Use available graphing software or tools to visually inspect this function.

Step 5: Identify the Maximum Area

To find the angle \( \theta \) that maximizes \( A(\theta) \), take the derivative of \( A(\theta) \) with respect to \( \theta \), and set it to zero. Solve \( \frac{d}{d\theta}(A(\theta)) = 0 \) to find potential maxima. Additionally, check endpoints and critical points within the interval \([0, \pi/2]\).

Key Concepts

Geometry of Rain Gutters

Rain gutters are essential for directing rainwater away from structures, protecting the property from water damage. A rain gutter's cross-sectional area determines how much water it can handle. Understanding the geometry of a rain gutter helps in designing efficient systems. In this exercise, we start with a metal sheet and transform it into a useful channel by bending up its sides.

When dealing with a rectangular sheet bent into a gutter:

• The central portion remains flat, serving as the base of the gutter.
• The side portions are bent upwards to form the gutter walls.
• This creates a structure similar to a tent, with two slanted sides meeting at the center base.

Visualizing this shape helps in calculating the cross-sectional area and later optimizing it for maximum efficiency. Understanding how the side angles impact the gutter's shape is crucial in maximizing the gutter's functionality.

Trigonometric Functions

Trigonometric functions play a vital role in this exercise by helping us describe the relationships between the angles and side lengths of the gutter. When bending the sides of a sheet, trigonometric functions like sine and cosine are used to describe the new dimensions.

Important points to remember include:

• \(\sin \theta\) represents the overlap of vertical height created for one side of the gutter when bent upwards.
• \(\cos \theta\) represents the effective horizontal reduction caused by the bending action, creating an altered base length for the gutter.

These functions are combined to derive the formula for the cross-sectional area as the gutter changes shape with varying angles \(\theta\). Mastering these functions helps us express complex geometric transformations in a manageable mathematical format.

Optimization of Area

Optimization is a vital mathematical process that helps in achieving the best possible outcome under given constraints. In this example, our goal is to find the angle \(\theta\) that maximizes the cross-sectional area of the gutter.

This involves:

• Deriving a function for the cross-sectional area that depends on \(\theta\).
• Analyzing how changes in \(\theta\) affect the area to find the maximum point.
• Utilizing the endpoints \(0\) and \(\pi/2\) in your evaluation to ensure you aren’t missing potential extremes.

The goal is to make the gutter capable of handling the largest possible volume of water, ensuring efficiency. Optimization provides the steps needed to refine and improve design performance through mathematical analysis.

Derivatives in Calculus

In calculus, derivatives allow us to understand how a function behaves by providing information about rates of change. Here, the derivative helps us find where the cross-sectional area of the gutter is at its maximum.

The steps include:

• Taking the derivative of the area function \(A(\theta)\) with respect to \(\theta\).
• Setting the derivative equal to zero to find critical points, which indicate potential maxima, minima, or points of inflection.
• Analyzing these points along with endpoints using a derivative test or second derivative test to confirm which yields the largest area.

This process identifies the optimal angle \(\theta\) by pinpointing the precise moment of maximal cross-sectional area. Derivatives simplify the search for optimization solutions in complex geometric scenarios.