Problem 426 - Calculus Volume 1

Question

For the following exercises, solve each problem. A chain hangs from two posts four meters apart to form a catenary described by the equation \(y=4 \cosh (x / 4)-3 .\) Find the total length of the catenary (arc length).

Step-by-Step Solution

Answer

The total length of the catenary is approximately 4.17 meters.

Step 1: Understand the Catenary Equation

The catenary is described by the equation \( y = 4 \cosh(x/4) - 3 \). The function \( \cosh(x) = \frac{e^x + e^{-x}}{2} \) is the hyperbolic cosine function. Thus, the function describes the shape of a hanging chain or cable.

Step 2: Set up the Arc Length Formula

The arc length \( L \) of a curve described by \( y=f(x) \) from \( x=a \) to \( x=b \) is given by the formula: \[ L = \int_a^b \sqrt{1 + \left(\frac{dy}{dx}\right)^2} \, dx \]

Step 3: Differentiate the Catenary Equation

Differentiate \( y = 4 \cosh(x/4) - 3 \) with respect to \( x \). \( \frac{dy}{dx} = \frac{d}{dx}(4 \cosh(x/4)) = 4 \sinh(x/4) \cdot \frac{1}{4} = \sinh(x/4) \)

Step 4: Substitute the Derivative into the Arc Length Formula

Substitute \( \frac{dy}{dx} = \sinh(x/4) \) into the arc length formula: \( L = \int_a^b \sqrt{1 + (\sinh(x/4))^2} \, dx \)

Step 5: Simplify the Integrand

Using the identity \( \cosh^2(u) - \sinh^2(u) = 1 \), we have \( 1 + \sinh^2(x/4) = \cosh^2(x/4) \). Thus, the integrand becomes \( \sqrt{\cosh^2(x/4)} = |\cosh(x/4)| = \cosh(x/4) \) since \( \cosh \) is always positive.

Step 6: Evaluate the Integral

Evaluate the integral: \( L = \int_{-2}^{2} \cosh(x/4) \, dx \)This becomes: \[ L = 4 \sinh(x/4) \Bigg|_{-2}^2 = 4 \left[ \sinh(1/2) - \sinh(-1/2) \right] \]Since \( \sinh(-u) = -\sinh(u) \), the expression simplifies to: \[ 4 \left[ \sinh(1/2) + \sinh(1/2) \right] = 8 \sinh(1/2) \]

Step 7: Calculate the Final Answer

Calculate \( \sinh(1/2) \), which is approximately 0.5211, so: \( L = 8 \times 0.5211 \approx 4.1688 \) meters.

Key Concepts

Catenary

The term "catenary" refers to the curve formed by a hanging flexible chain or cable when it's only supported at its ends. This shape is not a parabola, as one might intuitively expect, but is instead described by a hyperbolic cosine function. In the context of our problem, the equation given is \(y = 4 \cosh(x/4) - 3\). This becomes the mathematical representation of the catenary.

Some properties of the catenary include:

• It is symmetric with respect to the vertical axis that passes through its lowest point.
• Compared to a parabola, the dip of a catenary tends to be shallower.

The shape is quite useful in architecture and engineering, particularly for designing bridges and arches, due to its ability to evenly distribute weight.

Hyperbolic Cosine

The hyperbolic cosine function, denoted as \(\cosh(x)\), is essential to understanding catenaries. It is defined by the formula \(\cosh(x) = \frac{e^x + e^{-x}}{2}\). Unlike the traditional cosine function, \(\cosh(x)\) describes the shape of a hyperbola in a unit hyperbola form.

Some important characteristics of the \(\cosh(x)\) function include:

• Even function: \(\cosh(-x) = \cosh(x)\).
• Its graph has a U-like shape but does not oscillate like \(\cos(x)\).
• The minimum value is 1, which occurs at \(x = 0\).

In the problem, \(4 \cosh(x/4) - 3\) represents the form of the catenary equation, transforming the hyperbolic cosine to fit the specific constraints.

Integration

Integration is a fundamental concept in calculus used to find areas under a curve, among other applications. In geometry, especially for exercises involving curves, integration allows us to determine quantities like arc length.

For our arc length problem, we utilize the formula:\[ L = \int_a^b \sqrt{1 + \left(\frac{dy}{dx}\right)^2} \, dx \]Here, the integration process involves solving this integral over a defined interval, allowing us to accumulate the small pieces of the curve's length.

In our catenary problem, the integration bounds are from \(-2\) to \(2\) and the integrand simplifies with knowledge from trigonometry identities to easier functions like \(\cosh(x/4)\). This simplification is crucial as it makes the integration process straightforward.

Differentiation

Differentiation is the process of finding the derivative which measures how a function changes as its input changes. In the context of our catenary problem, finding the derivative \(\frac{dy}{dx}\) is a necessary step to calculate the arc length.

The given catenary equation is \( y = 4 \cosh(x/4) - 3\). The differentiation of this function yields:\[ \frac{dy}{dx} = \frac{d}{dx}(4 \cosh(x/4)) = 4 \sinh(x/4) \times \frac{1}{4} = \sinh(x/4) \]This result stems from the property of hyperbolic functions, where the derivative of \(\cosh(u)\) is \(\sinh(u)\).

Differentiation here is instrumental because it helps adjust the arc length formula, preparing it for successful integration.