Problem 64 - Calculus for Biology and Medicine

Question

A cable that hangs between two poles at \(x=-M\) and \(x=M\) takes the shape of a catenary, with equation $$ y=\frac{1}{2 a}\left(e^{a x}+e^{-a x}\right) $$ where \(a\) is a positive constant. Compute the length of the cable when \(a=1\) and \(M=\ln 2\).

Step-by-Step Solution

Answer

The length of the cable is \( \frac{3}{2} \).

Step 1: Set Up the Length Formula

To find the length of the cable, we use the general formula for the arc length of a function from \( x = -M \) to \( x = M \): \[l = \int_{-M}^{M} \sqrt{1 + \left( \frac{dy}{dx} \right)^2} \, dx. \] We will first need to find the derivative \( \frac{dy}{dx} \).

Step 2: Calculate the Derivative

The given equation is \[y = \frac{1}{2} (e^x + e^{-x}) \]Differentiating this equation with respect to \( x \), we get \[\frac{dy}{dx} = \frac{1}{2} (e^x - e^{-x}). \] This simplifies to the expression for the derivative.

Step 3: Simplify the Arc Length Integrand Formula

Now, substitute \( \frac{dy}{dx} \) into the arc length formula: \[l = \int_{-\ln 2}^{\ln 2} \sqrt{1 + \left(\frac{1}{2}(e^x - e^{-x})\right)^2} \, dx. \] Simplify the expression under the square root: \[1 + \left(\frac{1}{2}(e^x - e^{-x})\right)^2 = 1 + \frac{1}{4} (e^{2x} - 2 + e^{-2x}) = \frac{1}{2} (e^{2x} + e^{-2x})\]\[ = \frac{1}{2}(e^{2x} + 2 + e^{-2x} - 2) = \frac{1}{2}(e^{2x} + e^{-2x} - 2 + 2) = \frac{1}{2} \cdot \frac{1}{2}(e^{2x} + e^{-2x} + 2). \] This simplifies to the expression \[= \cosh^2(x).\] Therefore, the integrand becomes \( \cosh(x) \).

Step 4: Compute the Integral

The length of the cable is now \[l = \int_{-\ln 2}^{\ln 2} \cosh(x) \, dx. \] The antiderivative of \( \cosh(x) \) is \( \sinh(x) \). Thus we have:\[l = \left[ \sinh(x) \right]_{-\ln 2}^{\ln 2}. \]Evaluate this to get:\[l = \sinh(\ln 2) - \sinh(-\ln 2).\] Since \( \sinh(-x) = -\sinh(x) \), it simplifies to:\[l = 2\sinh(\ln 2). \]

Step 5: Calculate the Sinh Value

Now calculate \( \sinh(\ln 2) \): \[\sinh(\ln 2) = \frac{e^{\ln 2} - e^{-\ln 2}}{2} = \frac{2 - \frac{1}{2}}{2} = \frac{3}{4}. \]Thus, \[l = 2 \times \frac{3}{4} = \frac{3}{2}.\]This is the length of the cable.

Key Concepts

Catenary Equation

A catenary is the curve that a hanging flexible cable or chain assumes under its own weight when supported at its ends. Its shape is governed by the mathematical catenary equation. In our exercise, the equation is given by:

• \[ y = \frac{1}{2 a}(e^{ax} + e^{-ax}) \]

Here, \( a \) is a constant that influences the steepness and "tightness" of the curve. The larger the value of \( a \), the steeper the curve appears. In real-world scenarios, this equation helps engineers design suspension bridges and cables. For students learning calculus, understanding the catenary equation is crucial for solving problems involving these kinds of curves. Breaking down the equation, we see it comprises exponential functions with positive and negative exponents, providing the symmetrical shape of the curve. This foundational understanding paves the way for further exploration into hyperbolic functions and calculus applications.

Hyperbolic Functions

Hyperbolic functions, such as hyperbolic cosine (\(\cosh(x)\)) and hyperbolic sine (\(\sinh(x)\)), play a vital role in describing the catenary.

• Unlike circular trigonometric functions, hyperbolic functions relate to a hyperbola, much like how sine and cosine relate to a circle.


• They are used in various fields of mathematics and engineering due to their unique properties.

In the arc length problem, we encountered the hyperbolic cosine when simplifying the integrand under the square root. Using the identity:

• \[ \cosh^2(x) - \sinh^2(x) = 1 \]

we realize that \(\cosh(x)\) perfectly squares to the expression needing integration, simplifying our task significantly. Understanding these functions can help in identifying patterns and solutions in calculus problems that might otherwise seem complex.

Arc Length Formula

To find the length of a curve from one point to another, we utilize the arc length formula in integral calculus:

• \[ l = \int_{a}^{b} \sqrt{1 + \left( \frac{dy}{dx} \right)^2} \, dx \]

This formula provides a way to calculate the "smooth" length of the curve instead of the straight-line distance. The exercise involves determining this length between two points on a catenary. Steps in using this formula include:

• Finding the derivative of the function \( y \), which helps illustrate how the slope changes.


• Substituting this derivative back into the formula to set up the integral.

Understanding the arc length formula is crucial for determining the actual distance along any curved path, which is applicable in physics, engineering, and computer graphics.

Integral Calculus

Integral calculus is fundamental in solving the arc length problem since it allows us to calculate areas under curves, volumes, and as shown, lengths of intervals. The exercise uses the definite integral to find the arc length of the catenary from \(-\ln 2\) to \(\ln 2\).

• The integral setup is simplified using hyperbolic functions, allowing for easier computation. In our case, the integral of \( \cosh(x) \) yields \( \sinh(x) \), as these functions are intrinsically linked.


• After substituting in the limits, the evaluation gives the cable's length.

Integral calculus not only helps in measuring lengths and areas but also provides the mathematical groundwork for more advanced scientific and engineering applications. It breaks down complex problems into manageable calculations, often involving the replacement of complex curves with functions that are easier to integrate.