Problem 15 - Calculus: Early Transcendentals

Question

Find the arc length of the function on the given interval. $$ f(x)=\cosh x \text { on }[-\ln 2, \ln 2] $$

Step-by-Step Solution

Answer

The arc length is \( \frac{3}{2} \).

Step 1: Identify the Arc Length Formula

The formula to find the arc length of the function \( f(x) \) on the interval \([a, b]\) is given by \[ L = \int_a^b \sqrt{1 + [f'(x)]^2} \, dx \]. Here, \( f'(x) \) is the derivative of \( f(x) \).

Step 2: Derive the Function \( f(x) \)

Given \( f(x) = \cosh x \). The derivative \( f'(x) \) is calculated using the derivative of the hyperbolic cosine function, which is \( \sinh x \). So, \( f'(x) = \sinh x \).

Step 3: Substitute the Derivative in the Arc Length Formula

Substitute \( f'(x) = \sinh x \) into the arc length formula: \[ L = \int_{-\ln 2}^{\ln 2} \sqrt{1 + (\sinh x)^2} \, dx \].

Step 4: Simplify the Expression Under the Square Root

Notice that \( \cosh^2 x - \sinh^2 x = 1 \). Therefore, \( 1 + \sinh^2 x = \cosh^2 x \). Simplify the integrand: \[ L = \int_{-\ln 2}^{\ln 2} \sqrt{\cosh^2 x} \, dx = \int_{-\ln 2}^{\ln 2} \cosh x \, dx \].

Step 5: Integrate the Expression

Integrate \( L = \int_{-\ln 2}^{\ln 2} \cosh x \, dx \). The indefinite integral of \( \cosh x \) is \( \sinh x \). So, perform the definite integration: \[ L = \left[ \sinh x \right]_{-\ln 2}^{\ln 2} \].

Step 6: Evaluate the Definite Integral

Substitute the limits of integration: \( L = \sinh(\ln 2) - \sinh(-\ln 2) \). Note that \( \sinh(-x) = -\sinh(x) \). Therefore, this simplifies to \( L = \sinh(\ln 2) + \sinh(\ln 2) = 2\sinh(\ln 2) \).

Step 7: Calculate \( \sinh(\ln 2) \)

Recall the identity \( \sinh(a) = \frac{e^a - e^{-a}}{2} \). Substitute \( a = \ln 2 \) to get \( \sinh(\ln 2) = \frac{e^{\ln 2} - e^{-\ln 2}}{2} \). Calculate this to find \( \sinh(\ln 2) = \frac{2 - \frac{1}{2}}{2} = \frac{3}{4} \).

Step 8: Find the Final Arc Length Value

Multiply the result from Step 7 by 2: \( L = 2 \times \frac{3}{4} = \frac{3}{2} \). Hence, the arc length of the function \( f(x) = \cosh x \) on the interval \([-\ln 2, \ln 2]\) is \( \frac{3}{2} \).

Key Concepts

Hyperbolic Functions

Hyperbolic functions, much like their trigonometric counterparts, are essential in many areas of mathematics. These functions include hyperbolic sine (\(\sinh x\)), hyperbolic cosine (\(\cosh x\)), hyperbolic tangent, and others. They provide a bridge between exponential functions and geometric applications, often appearing in calculations involving hyperbolic angles and areas.

The hyperbolic cosine function, \(\cosh x\), has some interesting properties. For instance, it satisfies the identity \(\cosh^2 x - \sinh^2 x = 1\), similar to the trigonometric identity \(\cos^2 x + \sin^2 x = 1\). In this particular exercise, this identity is crucial as it helps simplify expressions leading to the computation of derivatives and integrals.

Definite Integral

The definite integral is a fundamental concept in calculus used to compute the accumulation of quantities. It is represented by the symbol \(\int_a^b f(x) \, dx\), where \(a\) and \(b\) are the limits of integration, and \(f(x)\) is the function being integrated.

In the context of finding arc length, the definite integral allows us to sum an infinite number of infinitesimally small sections of length along the curve. This is made possible by evaluating the integral from the start to the end of the interval, effectively computing the whole curve's arc length between two points. Understanding the integration process is key to solving problems involving continuous summation, such as this exercise.

Derivative Calculation

Derivatives measure how a function changes as its input changes. The process of finding a derivative is called differentiation, and it is a central operation in calculus.

For hyperbolic functions, like \(f(x) = \cosh x\), the derivative is found using the known rule for hyperbolic cosine, which states that \(f'(x) = \sinh x\). Calculating this derivative is essential in arc length problems because the arc length formula involves the derivative's square.

• Identify the function and its derivative formula.
• Calculate the derivative carefully to avoid errors.
• Substitute into the arc length formula and simplify if possible.

Finding derivatives and integrating them over intervals is a frequent task in calculus, making this concept both practical and theoretical.

Integration Techniques

Integration techniques are methods used to evaluate integrals, especially those that are not easily solved by straightforward antiderivatives. In the exercise, one can see the effective use of simplification techniques to make integration approachable.

For instance, transforming the integrand \( \sqrt{1 + (\sinh x)^2} \) to simply \(\cosh x\) by using the identity \(\cosh^2 x = 1 + \sinh^2 x\) dramatically simplifies the integration process. This simplification turns the integral into a more familiar form that is easier to evaluate. Such techniques are crucial as they reduce computation complexity and allow solutions to be more intuitive and less error-prone. Practicing these techniques will lead to greater flexibility in handling various integral problems.