Problem 64 - 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 largest cross-sectional area is achieved at \(\theta = \frac{\pi}{3}\).

Step 1: Define the Gutter in Terms of Geometry

The gutter is made by bending up one-third of the sheet on each side. Since the total width of the sheet is 30 cm, each side bent up accounts for \(\frac{30}{3} = 10\, \text{cm}\). When bent up at an angle \(\theta\), the height of each side is \(10 \sin \theta\), creating the configuration of a trapezoidal shape when viewed in cross-section.

Step 2: Express the Cross-Sectional Area

The gutter's cross-section consists of a central rectangle of width \(10 \cos \theta\) (since the base is halfway between the tops of the bent edges) and height \(10 \sin \theta\). Thus, the area is twice the triangle area plus the rectangle: \(A(\theta) = (10 \sin \theta)(10 \cos \theta) + 10 \sin \theta \cdot 10 \cos \theta\). Simplifying, this gives: \[A(\theta) = 100 \sin \theta + 100 \sin \theta \cos \theta.\]

Step 3: Graph the Function

To graph \(A(\theta) = 100 \sin \theta + 100 \sin \theta \cos \theta\), set up a coordinate system for \(\theta\) ranging from 0 to \(\frac{\pi}{2}\). The graph will show how the area changes as \(\theta\) increases. As \(\theta\) approaches \(\frac{\pi}{2}\), \(\cos \theta\) becomes smaller, affecting the second term.

Step 4: Find the Maximum Area

To find the maximum area, we need to maximize the function \(A(\theta) = 100 \sin \theta + 100 \sin \theta \cos \theta\). Calculate the derivative\[A'(\theta) = 100 \cos \theta + 100 \cos(2\theta),\] and set it to zero to find critical points. Solve \(100 (\cos \theta + \cos(2\theta)) = 0\). Simplifying using trigonometric identities, solve \(\cos \theta(1 + 2 \cos \theta) = 0\). The solutions, relevant to our domain, are \(\theta = \frac{\pi}{3}\).

Step 5: Confirm the Maximum Value

Check values slightly less and greater than \(\frac{\pi}{3}\) to confirm that the area decreases away from this point, ensuring \(\theta = \frac{\pi}{3}\) indeed gives a maximum in the range \(0\) to \(\frac{\pi}{2}\).

Key Concepts

Trigonometric Functions

When working with rain gutter geometry, trigonometric functions are crucial for analyzing angles and dimensions. In this exercise, the angle \( \theta \) plays a significant role because it determines the shape and area of the rain gutter's cross-section. Trigonometric functions like \( \sin \theta \) and \( \cos \theta \) describe the height and base of the triangular sections formed by bending the sheet. For instance:

• \( \sin \theta \) determines the vertical component or height of the triangles on either side of the gutter.
• \( \cos \theta \) is used to calculate the horizontal component, impacting both the position of the gutter's base and the width of its rectangular section.

Understanding these trigonometric functions helps us model the gutter's physical structure and find the optimal angle for maximum cross-sectional area.

Cross-Sectional Area

The cross-sectional area of a rain gutter is a geometric representation of the space within its boundaries, viewed from a direction perpendicular to its length. It can be thought of as the surface area that rain would touch if it flowed directly down into the gutter. In this problem, the area is influenced by both the rectangle and the two triangles formed by bending the edges:

• The rectangular part of the gutter has a width of \(10 \cos \theta\) and a height of \(10 \sin \theta\).
• Each triangular part contributes equally to the total area, thus multiplying by two.

The formula derived is \[A(\theta) = 100 \sin \theta + 100 \sin \theta \cos \theta\], encompassing the contribution from both the rectangle and triangles. This mathematical model helps visualize how the bent angles affect the rain gutter's capacity.

Derivative

The derivative of a function essentially measures how the function's output changes in response to changes in input. Here, it allows us to find the rate at which the cross-sectional area of the rain gutter changes as the angle \( \theta \) varies. By calculating \( A'(\theta) \), our derivative is:\[ A'(\theta) = 100 \cos \theta + 100 \cos(2\theta) \]Taking this derivative helps determine critical points where the function might achieve a maximum value.Calculating derivatives is a common method to solve optimization problems because it provides the mathematical tools to analyze and identify these crucial turning points.

Maximization Problem

The task of finding the maximum cross-sectional area is referred to as a maximization problem. Here, the goal is to determine the angle \( \theta \) at which the rain gutter can hold the most water. To solve this:

• We typically set the derivative \( A'(\theta) \) equal to zero to find critical points where the slope of the tangent to the curve is zero, indicating potential maxima or minima.
• In this exercise: \( 100 (\cos \theta + \cos(2\theta)) = 0 \), which simplifies to finding \( \cos \theta (1 + 2 \cos \theta) = 0 \).
• The solution is \( \theta = \frac{\pi}{3} \), showing the angle where the largest area occurs within the given domain of \( 0 \leq \theta \leq \frac{\pi}{2} \).
Testing values around this point confirms the nature of the maximum, ensuring the gutter's design can efficiently manage water flow.