Problem 70 - Algebra and Trigonometry

Question

Turning a Corner A steel pipe is being carried down a hallway that is 9 \(\mathrm{ft}\) wide. At the end of the hall there is a right-angled turn into a narrower hallway 6 \(\mathrm{ft}\) 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 is the longest pipe fitting the turn.

Step 1: Understanding the Geometric Scenario

Imagine a hallway where a right turn is needed. The problem involves a pipe being carried into another narrower hallway. To navigate this, we need to express the pipe's length in terms of an angle \( \theta \) that helps map out the geometry of turning the corner.

Step 2: Modeling the Pipe's Length

The length of the pipe \( L \) is influenced by two components: it spans across a 9 ft wide hallway and turns into the 6 ft narrow hallway. The equations \( 9 \csc(\theta) \) and \( 6 \sec(\theta) \) represent these parts where \( \csc(\theta) = \frac{1}{\sin(\theta)} \) manages the initial hallway width and \( \sec(\theta) = \frac{1}{\cos(\theta)} \) handles the turn into the narrower hallway.

Step 3: Graphing the Function

To visualize \( L(\theta) = 9 \csc(\theta) + 6 \sec(\theta) \), plot the graph for \( 0 < \theta < \frac{\pi}{2} \). The graph helps to understand how the length varies as the angle changes, focusing on finding the minimum.

Step 4: Differentiating to Find Critical Points

Calculate the derivative of \( L(\theta) \), which is \( L'(\theta) = -9 \csc(\theta) \cot(\theta) + 6 \sec(\theta) \tan(\theta) \). Set \( L'(\theta) = 0 \) to find critical points where the length has potential minimums or maximums.

Step 5: Solving for \( \theta \) when \( L'(\theta) = 0 \)

Use trigonometric identities and set the derivative equation to zero: \(-9 \csc(\theta) \cot(\theta) + 6 \sec(\theta) \tan(\theta) = 0\). Solving yields \( \theta = \tan^{-1} \left(\frac{2}{3}\right) \).

Step 6: Calculating the Minimum Length

Substitute the critical angle \( \theta = \tan^{-1} \left(\frac{2}{3}\right) \) back into the original function \( L(\theta) \) to find the minimum length. This calculation gives the longest possible pipe that can navigate the corner.

Step 7: Explaining the Result

This minimum value represents the longest pipe that can make the turn because it accounts for maximizing the pipe while adhering to hallway constraints. Thus, changes in \( \theta \) articulate the efficiency of space within the tight turn.

Key Concepts

Geometric Modeling

Geometric modeling forms the foundation of solving real-world problems by using shapes and angles to create a mathematical representation. In this exercise, we are asked to determine the length of a steel pipe that can turn a corner between two hallways using geometric principles. The key here is to think about how the pipe interacts with the walls as it twists to make a turn. Geometrically, we define an angle \( \theta \) that varies based on the pipe's orientation relative to the corner. This angle helps to map the two main segments of the pipe: one in the broader section of the hallway and one in the narrower section. We express the length of the pipe as a function, \( L(\theta) = 9 \csc(\theta) + 6 \sec(\theta) \). By decomposing this function:

• \( 9 \csc(\theta) \) represents the section of the pipe in the 9 ft wide hallway.
• \( 6 \sec(\theta) \) accounts for the pipe's path in the 6 ft wide hallway.

By modeling the scenario geometrically, we bridge the physical setup with a mathematical formula that expresses the full length of the pipe as it rounds the corner.

Optimization Problems

This exercise represents a classic optimization problem, where we determine the longest possible pipe that can maneuver around a corner. The goal is to find the minimum value of the length function \( L(\theta) \), as it depicts the longest allowable configuration of the pipe to successfully make the turn. Optimization involves analyzing how the function changes with respect to \( \theta \), the angle of turning. The method uses calculus to analyze trends and identify values that lead to optimal results. Through this problem, students learn how to apply optimization to real-world scenarios, ensuring efficient use of space. Some helpful steps include:

• Setting up a function that models the scenario, such as \( L(\theta) \).
• Analyzing the function graphically for potential maximum or minimum values.
• Calculating derivatives to find critical points, which reveal possible optimization solutions.

Derivatives in Calculus

Derivatives are a powerful tool in calculus used to analyze functions and determine rates of change. In optimization problems, derivatives help identify critical points like maximums and minimums, crucial for finding optimal solutions. For the pipe problem, we calculate the derivative of the function \( L(\theta) = 9 \csc(\theta) + 6 \sec(\theta) \): \[ L'(\theta) = -9 \csc(\theta) \cot(\theta) + 6 \sec(\theta) \tan(\theta) \] By setting this derivative equal to zero, we can find the critical points which in turn highlight where the function takes on minimum values. Solving \( L'(\theta) = 0 \) brings us to the critical angle \( \theta = \tan^{-1} \left(\frac{2}{3}\right) \), aligning with optimizing the function length. Derivatives not only guide finding key points in functions but also provide insight into how small changes in \( \theta \) affect the overall outcome, illuminating the delicate balance needed to successfully navigate the corner with the longest possible pipe.