Problem 41 - Calculus of a Single Variable

Question

In Exercises 41 and 42, a model for a power cable suspended between two towers is given. (a) Graph the model, (b) find the heights of the cable at the towers and at the midpoint between the towers, and (c) find the slope of the model at the point where the cable meets the right-hand tower. $$ y=10+15 \cosh \frac{x}{15}, \quad-15 \leq x \leq 15 $$

Step-by-Step Solution

Answer

The graph is a curve showing the cable suspended between two towers. The height of the cable at the towers is \(10 + 15 \cosh(1)\), and its height at the midpoint is \(25\). The slope of the cable at the right-hand tower is \(\sinh(1)\).

Step 1 - Graph the Model

Graph the function \( y = 10 + 15 \cosh(x/15) \) in the interval \(-15 \leq x \leq 15\). With modern technology, plotting the graph should be a straightforward task. The graph depicts a cable hanging between two towers, which are 30 units apart.

Step 2 - Find the Heights of the Cable

The cable's heights at the towers and the midpoint can be found by substituting these positions into the equation. At \( x = -15, 0, \) and \( 15 \), calculate \( y \). For \( x = -15 \) or \( x = 15 \), the result is \( y = 10 + 15 \cosh(1) \). For \( x = 0 \), \( y = 10 + 15 \cosh(0) = 25 \), because \( \cosh(0) = 1 \). This represents the highest and lowest points.

Step 3 - Find the Slope at Right-hand Tower

The slope of the model at any given point is found by taking the derivative of the function at that point. The derivative of \( \cosh \) is \( \sinh \). Hence, \( y' = 15 \sinh(x/15) / 15 = \sinh(x/15) \). Now, find the value at \( x = 15 \) which results in \( y'(15) = \sinh(1) \), which is the slope at the right-hand tower.

Key Concepts

Graphing Functions

Graphing functions involves representing a mathematical equation visually on a coordinate plane. For the given equation, we are interested in the function \( y = 10 + 15 \cosh(x/15) \) over the interval \( -15 \leq x \leq 15 \). The graph of this function appears as a catenary, resembling a U-shape that is symmetric around the y-axis.

• The graph reveals important insights, such as the height of the cable at different positions along the span.
• Using graphing tools or software, you can plot this equation to see how the cable spans between the two towers.
• The highest point, known as the vertex, occurs at the midpoint \(x = 0\).

This step doesn't just help visualize the problem but also sets a foundation for further analysis by showing the curvature and general shape.

Derivatives

Derivatives are fundamental in calculus, allowing us to determine the rate of change or slope of a function at any point. To find the derivative of our hyperbolic function \( y = 10 + 15 \cosh(x/15) \), we'll focus on the term involving the hyperbolic cosine.
The derivative of \( \cosh(u) \) is \( \sinh(u) \), so the chain rule helps us differentiate \( y = 10 + 15 \cosh(x/15) \) with respect to \(x\). The derivative \( y' \) becomes:

• \( y' = 15 \cdot \sinh(x/15) \cdot \frac{d}{dx}(x/15) = \sinh(x/15) \)

With derivatives, we find slopes of tangents at specific points and determine where functions increase, decrease, or have extrema.

Catenary Curve

A catenary curve is formed by a hanging cable or chain under its weight, assuming the shape described by a hyperbolic cosine function. This particular curve is represented by \( y = 10 + 15 \cosh(x/15) \).

• Unlike a parabola often found in traditional quadratic functions, the catenary captures realistic suspension curves due to gravity.
• It represents an equilibrium position where tensile forces in the cable balance perfectly.
• This is illustrated between two points of suspension, such as two towers.

Knowing the mathematical definition of a catenary helps in understanding engineering problems and applications involving arches, bridges, and suspended cables.

Chain Rule

The chain rule is a powerful tool in calculus used to differentiate composite functions. When dealing with the function \( y = 10 + 15 \cosh(x/15) \), we use the chain rule to find the derivative effectively.
Here's how it assists:

• The function is a composition of \( \cosh(u) \) where \( u = x/15 \), making it necessary to apply the chain rule.
• The outer function is \( \cosh(u) \), and the inner function is \( u = x/15 \).
• By differentiating \( \cosh(x/15) \) with respect to \( x \), and using \( \frac{d}{du}(\cosh(u)) = \sinh(u) \), we apply the chain rule as follows:

For the differential, we calculate:

• \( \frac{d}{dx}(10 + 15 \cosh(x/15)) = \sinh(x/15) \).

Using the chain rule allows us to tackle complex problems by breaking them down into simpler parts.