Problem 81 - Thomas Calculus

Question

Arc length Find the length of the graph of \(y=(1 / 2) \cosh 2 x\) from \(x=0\) to \(x=\ln \sqrt{5} .\)

Step-by-Step Solution

Answer

The arc length is 1.2.

Step 1: Understand the Problem and Formula for Arc Length

We need to find the arc length of the curve given by \(y = \frac{1}{2} \cosh{2x}\) from \(x = 0\) to \(x = \ln \sqrt{5}\). The formula for the arc length \(L\) for a function \(y = f(x)\) from \(x = a\) to \(x = b\) is given by:\[L = \int_a^b \sqrt{1 + \left(\frac{dy}{dx}\right)^2} \, dx\] where \(\frac{dy}{dx}\) is the derivative of \(y\).

Step 2: Differentiate the Function

Differentiate \(y = \frac{1}{2} \cosh{2x}\) with respect to \(x\):\[\frac{dy}{dx} = \frac{1}{2} \cdot 2 \sinh{2x} = \sinh{2x}\]

Step 3: Set Up the Arc Length Integral

Substitute \(\frac{dy}{dx}\) into the arc length formula:\[L = \int_0^{\ln \sqrt{5}} \sqrt{1 + (\sinh{2x})^2} \, dx\] Recall the identity: \(\cosh^2{u} - \sinh^2{u} = 1\). So, \(1 + \sinh^2{2x} = \cosh^2{2x}\).

Step 4: Simplify the Integral

The integral becomes:\[L = \int_0^{\ln \sqrt{5}} \sqrt{\cosh^2{2x}} \, dx = \int_0^{\ln \sqrt{5}} \cosh{2x} \, dx\]

Step 5: Evaluate the Integral

Evaluate \(\int \cosh{2x} \, dx\): The antiderivative of \(\cosh{2x}\) is \(\frac{1}{2} \sinh{2x}\). Evaluate from \(x = 0\) to \(x = \ln \sqrt{5}\):\[L = \left. \frac{1}{2} \sinh{2x} \right|_0^{\ln \sqrt{5}} = \frac{1}{2} \sinh(2 \ln \sqrt{5}) - \frac{1}{2} \sinh(0)\]

Step 6: Calculate the Result

First, simplify \(2 \ln \sqrt{5}\) as \(\ln 5\) (since \(2\cdot(\ln \sqrt{5}) = \ln (5^{1/2})^2 = \ln 5\)). Thus, compute \(\sinh(\ln 5)\):\[\sinh(\ln 5) = \frac{e^{\ln 5} - e^{-\ln 5}}{2} = \frac{5 - \frac{1}{5}}{2} = \frac{24}{10} = 2.4\]The arc length is \(L = \frac{1}{2} \times 2.4 = 1.2\).

Step 7: Final Result

The arc length of the graph of \(y = \frac{1}{2} \cosh 2x\) from \(x = 0\) to \(x = \ln \sqrt{5}\) is \(L = 1.2\).

Key Concepts

Hyperbolic Functions

The hyperbolic functions, much like trigonometric functions, are important in various branches of mathematics, especially in calculus and complex analysis. These functions include \(\sinh(x)\) and \(\cosh(x)\), which are vital to solving problems involving the arc length of curves. Hyperbolic functions are closely related to exponential functions, and they can be expressed as follows:

• \(\sinh(x) = \frac{e^x - e^{-x}}{2}\)
• \(\cosh(x) = \frac{e^x + e^{-x}}{2}\)

These functions exhibit unique properties:

• They are symmetric concerning the y-axis.
• The derivative of \(\sinh(x)\) is \(\cosh(x)\) and vice versa.

The identity \(\cosh^2(x) - \sinh^2(x) = 1\) is analogous to the Pythagorean identity for trigonometric functions, and it plays a crucial role in simplifying expressions involving hyperbolic functions.

Integration

Integration is a fundamental concept in calculus, used to find areas, volumes, and in this context, arc lengths. The process of integration involves finding an antiderivative or evaluating a definite integral over a specific interval. For the arc length of a curve, we use the arc length formula:

\[L = \int_a^b \sqrt{1 + \left(\frac{dy}{dx}\right)^2} \, dx\]
In this particular problem, we focused on evaluating \(\int_0^{\ln \sqrt{5}} \cosh(2x) \, dx\), using known identities and antiderivatives of hyperbolic functions.

Definite integrals involve evaluating the antiderivative at the upper limit and subtracting the value at the lower limit, providing the total arc length over the given interval.

Derivatives

Derivatives are instrumental in finding the rate of change of a function and are crucial for setting up the arc length formula. In this exercise, we started with the derivative of \(y = \frac{1}{2} \cosh{2x}\), calculated as follows:

\[\frac{dy}{dx} = \frac{1}{2} \cdot 2 \sinh{2x} = \sinh{2x}\]
Understanding derivatives helps in determining aspects like slopes and instantaneous rates of change. In this context, the derivative was used to form the function inside the arc length integral, leading to simplification using hyperbolic identities.

Cosh and Sinh Functions

The functions \(\cosh(x)\) and \(\sinh(x)\) are core hyperbolic functions commonly encountered in calculus, similar to cosine and sine in trigonometry. They have unique applications due to their exponential nature, and certain identities associated with these functions simplify integration and differentiation:

• The identity \(\cosh^2(u) - \sinh^2(u) = 1\) is used to simplify expressions in integrals.
• The integral of \(\cosh(2x)\) was determined as \(\frac{1}{2} \sinh(2x)\), because \(\frac{d}{dx} \sinh(x) = \cosh(x)\).

These simplifications allow us to solve problems involving arc lengths efficiently, showcasing the power of understanding hyperbolic functions in calculus.