Question

Assume a hanging cable has the shape \(15 \cosh (x / 15)\) for \(-20 \leq x \leq 20\). Determine the length of the cable (in feet).

Short answer

The length of the cable is approximately 61.01 feet.

Step-by-step solution

Understand the Parametric Equation

The given function for the cable is the hyperbolic cosine function \(15 \cosh(x / 15)\). We need to find the length of this curve from \(x = -20\) to \(x = 20\). The curve is described in parametric form using the hyperbolic cosine function, which is a standard curve similar to a hanging cable or chain, known as the catenary curve.

Set up the Arc Length Formula

The formula for finding the arc length \(L\) of a function \(y = f(x)\) from \(x = a\) to \(x = b\) is: \[L = \int_a^b \sqrt{1 + (f'(x))^2} \, dx\]. For this problem, \(f(x) = 15 \cosh(x / 15)\) and \(a = -20\), \(b = 20\).

Differentiate the Function

Find the derivative \(f'(x)\). Given that \(f(x) = 15 \cosh(x/15)\), we use the derivative formula for \(\cosh\), resulting in: \[f'(x) = \frac{d}{dx}[15 \cosh(x/15)] = 15 \sinh(x/15) \cdot \frac{1}{15} = \sinh(x/15)\].

Substitute derivative into Arc Length Formula

Now substitute \(f'(x)\) into the arc length formula: \[L = \int_{-20}^{20} \sqrt{1 + \sinh^2(x/15)} \, dx\]. Use the identity \(1 + \sinh^2(u) = \cosh^2(u)\) to simplify: \[L = \int_{-20}^{20} \cosh(x/15) \, dx\].

Integrate the Function

The integral \(\int \cosh(u) \, du = \sinh(u)\). So, \[L = \left[15 \sinh(x/15)\right]_{-20}^{20}\]. Evaluate this step-by-step: compute \(15 \sinh(20/15)\) and \(15 \sinh(-20/15)\).

Evaluate the Integral

Calculate \(\sinh(20/15)\) and \(\sinh(-20/15)\). Since \(\sinh\) is an odd function, \(\sinh(-u) = -\sinh(u)\). Compute \[L = 15 \sinh(20/15) - 15 (-\sinh(20/15)) = 30 \sinh(20/15)\].

Final Calculation

Finally, calculate the numerical value of \(30 \sinh(20/15)\). Use a calculator to find the approximate value of \(\sinh(20/15)\) and multiply by 30.

Key concepts

Hyperbolic Functions

Hyperbolic functions are mathematical functions similar to the trigonometric functions. However, instead of circles, these functions relate to the geometry of hyperbolas. Two main hyperbolic functions are the hyperbolic cosine, \(\cosh(x)\), and the hyperbolic sine, \(\sinh(x)\). These functions have exponential expressions:

• For \(\cosh(x)\), the formula is \(\frac{e^x + e^{-x}}{2}\).
• For \(\sinh(x)\), it is \(\frac{e^x - e^{-x}}{2}\).

Hyperbolic functions emerge in scenarios involving hyperbolas, such as the path of a hanging cable, due to their unique properties in stretching and tension forces. These functions are crucial for problems involving catenary curves, which naturally occur in physics and engineering.

Catenary Curve

A catenary curve is the shape that a hanging flexible chain or cable assumes under its weight when supported only at its ends. This specific curve is intimately linked with hyperbolic cosines, described by the equation \(y = a \cosh\left(\frac{x}{a}\right)\). The hyperbolic cosine function fits such a scenario neatly because of its properties.
The term 'catenary' comes from the Latin word 'catena,' meaning 'chain.' This shape minimizes potential energy, a principle that explains its natural occurrence in freely hanging cables.
A clear recognition of the catenary curve is key while tackling problems involving arc lengths of hyperbolic paths. It simplifies the modeling of real-world problems involving overhead cables or bridges.

Integral Calculus

Integral calculus is essential for determining quantities like areas under a curve or lengths of curves. This branch of calculus deals with integrals and their properties, stemming from the idea of summing infinitely small parts to find the whole.
Arc length problems, similar to the exercise concerning the catenary curve, require using definite integrals. For a curve \(y = f(x)\) from \(x = a\) to \(x = b\), the arc length \(L\) is found using the integral formula: \[L = \int_a^b \sqrt{1 + \left(\frac{dy}{dx}\right)^2} \, dx\].

• This formula arises from Pythagorean theorem concepts, where \(\sqrt{1 + (dy/dx)^2}\) represents the rate of change considering both horizontal and vertical components as you move along the curve.
• To solve for specific lengths, one needs to evaluate the definite integral through limits of the given function.

Integral calculus allows for precise calculations of such quantities crucial in both theoretical and applied mathematics.

Derivatives

Derivatives represent the rate at which a function changes at any given point, often described as the "slope" of the function. Calculating derivatives is a fundamental skill in calculus.
The derivative of the function \(y = 15 \cosh(x/15)\) in the catenary problem is derived using the standard derivative rule for hyperbolic functions. Here, \(\frac{d}{dx}[\cosh(u)] = \sinh(u)\frac{du}{dx}\). Since \(x\) is scaled by 15, using chain rule adjustments, the result is \(\sinh(x/15)\).
Derivatives are critical for determining the nature of change in arc length as well as understanding dynamic systems. With derivatives, we can understand the varying steepness of curves, helping in both constructing integrals for arc lengths and modifying functions to suit physical constraints.