Problem 68 - Algebra and Trigonometry

Question

Turning a Corner A steel pipe is being carried down a hallway 9 ft wide. At the end of the hall there is a right-angled turn into a narrower hallway 6 ft wide. (a) Show that the length of the pipe in the figure is modeled by the function $$L(\theta)=9 \csc \theta+6 \sec \theta$$ (b) Graph the function \(L\) for \(0<\theta<\pi / 2\) (c) Find the minimum value of the function \(L\) (d) Explain why the value of \(L\) you found in part (c) is the length of the longest pipe that can be carried around the corner.

Step-by-Step Solution

Answer

The minimum length of the pipe determines the longest possible pipe that can fit around the corner without bending.

Step 1: Understand the Problem

A steel pipe is being carried down a hallway that turns a corner. The wider hallway is 9 ft wide and the narrower one is 6 ft wide. We are given a formula to model the length of such a pipe as a function of the angle \( \theta \) at the corner.

Step 2: Derive the Function

Recognize that the function \( L(\theta) = 9 \csc \theta + 6 \sec \theta \) represents the length of the pipe in terms of \(\theta\), the angle formed between the pipe and the narrow hallway after turning the corner. This formula uses trigonometric identities: \( \csc \theta = \frac{1}{\sin \theta} \) and \( \sec \theta = \frac{1}{\cos \theta} \), reflecting the geometry at the corner with one segment projecting across the width of the 9 ft hallway and the other along the narrower 6 ft hallway.

Step 3: Graph the Function

To graph \( L(\theta) = 9 \csc \theta + 6 \sec \theta \), plot the angle \( \theta \) on the x-axis and \( L(\theta) \) on the y-axis for the domain \( 0 < \theta < \frac{\pi}{2} \). Note the behavior where \( \theta \) approaches 0 or \( \frac{\pi}{2} \), highlighting vertical asymptotes.

Step 4: Calculate Minimum Length

To find the minimum value of \( L(\theta) \), differentiate \( L(\theta) \) with respect to \( \theta \):\[ L'(\theta) = -9 \csc \theta \cot \theta + 6 \sec \theta \tan \theta \]Set the derivative \( L'(\theta) = 0 \) and solve for \( \theta \) to determine critical points. Evaluate \( L(\theta) \) at these points and any end behavior if necessary to find the minimum length.

Step 5: Validate Longest Pipe Explanation

The minimum value of \( L(\theta) \) represents the longest pipe that can be carried around the corner because as \( \theta \) varies, the pipe's interaction with the corner is constrained by the hallway widths. The minimum point ensures maximum accommodation within these geometric restrictions.

Key Concepts

Trigonometric Functions

Trigonometric functions are mathematical functions of an angle common in various branches of science and engineering, including solving problems involving pipes around corners. These functions relate the angles of a triangle to the lengths of its sides. Some primary trigonometric functions include sine ( \( \sin \theta \) ), cosine ( \( \cos \theta \) ), and their reciprocals; cosecant ( \( \csc \theta = \frac{1}{\sin \theta} \) ) and secant ( \( \sec \theta = \frac{1}{\cos \theta} \) ).
These functions help model real-world shapes and angles, such as the steel pipe turning around the corner in the hallway. Understanding how these functions behave, such as tending towards infinity near certain angles, is crucial for applying them in practical situations like the one illustrated in the exercise.

Critical Points

Critical points are where the derivative of a function is zero or undefined. They help to find where a function reaches its maximum or minimum values.
In the context of the exercise, critical points are used to determine the length of the steel pipe that can be carried around a corner in the hallways. By identifying where the derivative of the pipe length function, \( L'(\theta) \) , is zero, we can locate potential minima or maxima.
That location can inform us of the longest possible length of pipe that fits through the hallway corner in the most efficient manner.

• Set the derivative \( L'(\theta) = -9 \csc \theta \cot \theta + 6 \sec \theta \tan \theta \) to zero.
• Solve for \( \theta \) to find critical points.
• Evaluate the function at these points and potentially check endpoints to ensure finding true minimum is necessary.

Derivative of Trigonometric Functions

Derivatives give us crucial information about the behavior of functions, such as rates of change, maxima, and minima. With trigonometric functions, understanding their derivatives is key to solving optimization problems like the hallway pipe problem.
For functions involving \( \csc \theta \) and \( \sec \theta \) , their respective derivatives are often tricky, requiring careful attention to trigonometric identities:

• Derivative of \( \csc \theta: -\csc \theta \cot \theta \) .
• Derivative of \( \sec \theta: \sec \theta \tan \theta \) .

The derivative of the length function, \( L'(\theta) = -9 \csc \theta \cot \theta + 6 \sec \theta \tan \theta \) , combines these concepts to find where the slope is zero, thus identifying critical points corresponding to the maximum possible length of the pipe fitting in the corner.

Geometry of Angles

The geometry of angles is foundational to understanding and solving trigonometric problems. The problem of the pipe in the hallway emphasizes the importance of angle geometry. When the pipe goes around the corner, an angle \( \theta \) forms between the pipe and the hallway.
This angle greatly impacts the possible maximum length of the pipe that can fit. By adjusting \( \theta \) , you see variations in the achievable pipe length due to the geometric restrictions defined by the widths of the hallways.
Understanding this geometric setup allows you to model it with trigonometric functions accurately and facilitates finding the most compact route for the pipe around the corner. This relationship between the angle and hallway dimensions elucidates the design and optimization process, linking the theoretical with the practical.