Problem 38 - Calculus Early Transcendentals: Pearson New International Edition

Question

Call the graph of \(y=b-a \cosh (x / a)\) an inverted catenary and imagine it to be an arch sitting on the \(x\) -axis. Show that if the width of this arch along the \(x\) -axis is \(2 a\) then each of the following is true. (a) \(b=a \cosh 1 \approx 1.54308 a\). (b) The height of the arch is approximately \(0.54308 a\). (c) The height of an arch of width 48 is approximately \(13 .\)

Step-by-Step Solution

Answer

(a) \(b = 1.54308a\). (b) Height = \(0.54308a\). (c) Height for width 48 is approximately 13.

Step 1: Understand the equation of the arch

The given equation is for an inverted catenary: \( y = b - a \cosh(x/a) \). The arch is centered on the x-axis, and its width is given as \(2a\). Here, \(b\) is the vertical height from the x-axis to the highest point of the arch.

Step 2: Find the endpoints of the arch on the x-axis

Since the width of the arch on the x-axis is \(2a\), the endpoints on the x-axis are at \(x = -a\) and \(x = a\). The graph of the inverted catenary is symmetric about the y-axis.

Step 3: Calculate the height of the arch at x=0

To find the height at the vertex of the arch, substitute \(x = 0\) into the equation: \[y = b - a \cosh(0/a) = b - a \cdot 1 = b - a.\]At \(x = 0\), the height is \(b - a\).

Step 4: Use the endpoint condition to solve for b

At the endpoints \(x = -a\) and \(x = a\), \(y = 0\). Substitute \(x = a\) into the equation:\[0 = b - a \cosh(a/a) = b - a \cosh(1).\]Solve for \(b\):\[ b = a \cosh(1).\]

Step 5: Approximate values for b and height

Using the identity \(\cosh(1) \approx 1.54308\), we have:\[ b = a \cdot 1.54308 = 1.54308a.\]The height of the arch at its peak (vertex) is:\[ Height = b - a = 1.54308a - a = 0.54308a.\]

Step 6: Calculate the height for an arch of width 48

Given the width of 48, solve for \(a\):\[ 2a = 48 \Rightarrow a = 24.\]Using the height calculation from the prior step, the height is:\[ 0.54308 \times 24 = 13.03392 \approx 13.\]

Key Concepts

Hyperbolic Functions

Hyperbolic functions are mathematical functions that share similarities with trigonometric functions, but instead of circular, they are based on hyperbolas. The most common hyperbolic functions are the hyperbolic sine (\( \sinh \)), hyperbolic cosine (\( \cosh \)), and hyperbolic tangent (\( \tanh \)). These functions are expressed using exponential functions and can be defined as follows:

• \( \sinh(x) = \frac{e^x - e^{-x}}{2} \)
• \( \cosh(x) = \frac{e^x + e^{-x}}{2} \)
• \( \tanh(x) = \frac{\sinh(x)}{\cosh(x)} \)

In the context of an inverted catenary, the hyperbolic cosine \( \cosh \) function plays a vital role. This function describes the shape of an idealized hanging chain or cable, commonly referred to as a catenary. When we invert this catenary to form an arch, \( \cosh \) helps determine both the height at different points of the arch and its symmetry across the y-axis.
The value of \( \cosh(1) \) is approximately 1.54308, which is crucial when calculating the specific parameters of the inverted catenary in the problem.

Arch Height

The height of an inverted catenary arch is a measure of its maximum vertical dimension above the x-axis at its center point. To find the arch height, we first need to determine the formula that gives this specific measurement. Given the standard equation of an inverted catenary: \[ y = b - a \cosh(x/a) \]The maximum height of the arch occurs at its center point, \( x=0 \). At this point, the equation simplifies to:\[ y = b - a \cosh(0) = b - a \cdot 1 = b - a \]Using the relationship derived from the problem, \( b = a \cosh(1) \), the rectangle calculates as:\[ b - a = a \cosh(1) - a = 1.54308a - a = 0.54308a \]This means the arch height is approximately 54.308% of the parameter \( a \). Understanding this relationship helps us determine the peak height of the arch whenever its width is known.

Catenary Width

The width of an inverted catenary arch is a fundamental measurement, defined along the x-axis from one end of the arch to the other. In the standard setup for our problem, the width of the arch is represented by \( 2a \). This is derived from the endpoints of the arch falling symmetrically on the x-axis at positions \( x = -a \) and \( x = a \).
The relationship \( 2a \) outlines the span across the x-axis and is crucial in designing structures that follow the principles of inverted catenaries, like bridges and arches. To illustrate, in the specific exercise, if the width is specified as 48, then solving for \( a \) yields:

• \( 2a = 48 \)
• \( a = \frac{48}{2} = 24 \)

Knowing \( a \) allows us to use it in calculating other important aspects like arch height. For instance, we found that for a width of 48, the corresponding arch height is approximately 13. This width-to-height relationship is key in practical applications, ensuring structures are both stable and aesthetically pleasing.