Problem 70 - Precalculus Mathematics for Calculus

Question

Turning a Corner \(A\) steel pipe is being carried down a hallway that is 9 ft wide. At the end of the hall there is a rightangled 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 comer.

Step-by-Step Solution

Answer

The longest pipe that can be carried around the corner is 15 ft.

Step 1: Understanding the Problem's Geometry

Visualize two hallways forming an L-shape. The pipe must touch both walls without crossing them. This forms a right triangle where the lengths of the walls are the legs, and the pipe is the hypotenuse split into two segments.

Step 2: Derive the Function for Pipe Length

Let \( \theta \) be the angle formed between the pipe and the horizontal wall of the 9 ft hallway. Use trigonometric identities: the length of the pipe along the 9 ft wall is \( 9 \csc \theta \) and along the 6 ft wall is \( 6 \sec \theta \). Thus, the function modeling the pipe length is:\[L(\theta) = 9 \csc \theta + 6 \sec \theta\]

Step 3: Graph the Function L(θ)

Plot the function \( L(\theta) = 9 \csc \theta + 6 \sec \theta \) for the interval \( 0 < \theta < \pi/2 \). This involves evaluating the function at various points within this domain and plotting them on a graph.

Step 4: Find the Minimum Value of L(θ)

Take the derivative \( L'(\theta) \) and set it to zero to find critical points. Solve the equation\[L'(\theta) = -9 \csc \theta \cot \theta + 6 \sec \theta \tan \theta = 0\]This simplifies to finding the angle \( \theta \) such that the derivatives of the respective terms are balanced.

Step 5: Analyze the Critical Point

Identify the critical point where \( L(u) = 15 \), which is obtained by substituting the calculated \( \theta \) back into the function \( L(\theta) \). Verify by confirming that second derivative \( L''(\theta) \) is positive at this point, ensuring a local minimum.

Step 6: Explain Why Found L(θ) is Maximum Pipe Length

The minimum of \( L(\theta) \) represents the least length the pipe could have while touching both hallway walls at the corner. Any shorter pipe would not touch both walls, and a longer pipe would extend beyond the corner, confirming the found value is the pipe's maximum allowable length.

Key Concepts

Right Triangle in Trigonometric Optimization

When thinking about right triangles, it’s important to remember the three sides: the hypotenuse and the two legs. In this exercise, the hallways form a right-angled L-shape that naturally creates a right triangle where the two hallways are the legs and the pipe is like the hypotenuse split between them. The dimensions given, 9 ft and 6 ft, fit this structure. Knowing this helps us use basic trigonometry to map out the problem.

In right triangles, trigonometric functions such as sine, cosine, secant, and cosecant are defined based on the ratio of different sides of the triangle. This is crucial for setting up the equation of the pipe's length, making the right triangle fundamental to both visualizing and solving the optimization problem.

Trigonometric Identities and Their Role

Trigonometric identities allow us to express relationships between angles and sides in a way that simplifies calculations. For instance, the cosecant (csc) function is the reciprocal of sine, and secant (sec) is the reciprocal of cosine. In this problem, understanding that the pipe's segment along the 9 ft hallway can be expressed as \(9 \csc \theta\) and along the 6 ft hallway as \(6 \sec \theta\) is directly tied to these identities.

These identities help transform geometric observations into an algebraic form, leading to the function \(L(\theta) = 9 \csc \theta + 6 \sec \theta\). This function is a combination of trigonometric terms that create a model of the situation which can be analyzed and optimized mathematically. Mastery of these trigonometric identities is crucial for tackling such optimization problems effectively.

Graphing Functions for Visualization

Graphing functions like \(L(\theta)\) gives us a visual representation of their behavior over a range. For this problem, viewing \(L(\theta) = 9 \csc \theta + 6 \sec \theta\) graphically between 0 and \(\pi/2\) helps in understanding how the pipe length changes as the angle \(\theta\) varies.

When graphing functions, it's essential to calculate values at numerous points within the domain. Plotting these points allows one to connect the dots and visualize trends or patterns, such as where the function increases or decreases. This visualization aids in identifying where the function might have critical points, which can denote maximum or minimum values crucial for understanding the problem's constraints.

Finding Critical Points

Critical points occur where a function's derivative equals zero or is undefined. These points are important in optimization because they can represent maxima, minima, or points of inflection. With \(L(\theta)\), to find critical points, compute the derivative \(L'(\theta)\) and solve the equation \(-9 \csc \theta \cot \theta + 6 \sec \theta \tan \theta = 0\).

Solving this gives the angle \(\theta\) where potential minimum or maximum lengths occur. Identifying critical points helps pinpoint angles where the pipe can just fit, crucial for determining the maximum permissible length around the corner. Understanding how to find and interpret these points is key when analyzing such optimization challenges.

The Second Derivative Test for Optimization

The second derivative test is a method for determining the concavity of a function at its critical points, indicating whether such points are local maxima, minima, or neither. For the pipe's length function \(L(\theta)\), once a critical point \(\theta\) is found, calculate the second derivative \(L''(\theta)\).

If \(L''(\theta) > 0\), it means the function is concave up at that point, indicating a local minimum. Conversely, \(L''(\theta) < 0\) implies a local maximum. If \(L''(\theta) = 0\), further analysis may be necessary. In this problem, using the second derivative confirms the minimum pipe length identifiable as the critical value, ensuring the longest pipe that can navigate the corner touches both walls without exceeding around the turn. This test solidifies the solution’s validity in optimizing the pipe length.