91Ó°ÊÓ

Differentiation is a fundamental concept in calculus where we find the rate at which a function is changing at any given point. To find the derivative, or the slope, of a function, we apply differentiation rules. In this exercise, the function given is hyperbolic: \[ y = \frac{1}{2} (e^x + e^{-x}) \]. To differentiate this function, we need the basic derivatives of exponential functions:

• The derivative of \( e^x \) is \( e^x \).
• The derivative of \( e^{-x} \) is \(-e^{-x} \).

Applying these, we find the derivative: \[ \frac{dy}{dx} = \frac{1}{2} (e^x - e^{-x}) \]. This derivative tells us how steep the curve is at any point \( x \). Differentiation is key in finding the arc length because we use the derivative in the arc length formula. Understanding how to calculate derivatives will help simplify complex integral calculations in such problems.

Hyperbolic functions are analogs of trigonometric functions but for the hyperbola. They often appear in calculus and include functions such as sinh, cosh, and tanh. For our function, \[ y = \frac{1}{2} (e^x + e^{-x}) \], which turns out to be related to the hyperbolic cosine function \( \cosh(x) \). What makes hyperbolic functions interesting is their close relation to exponential functions. For example:

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

This means our function is simply \( y = \cosh(x) \). Recognizing this allows simplifications in calculus problems, such as finding derivatives or solving integrals that involve arc lengths, because hyperbolic identities can make complex square root expressions simpler. This familiarity can significantly ease problem-solving and deepen understanding.

Integral calculus is used to find quantities like areas, volumes, and lengths, which are represented as integrals. For arc length, we use the integral formula: \[ L = \int_{a}^{b} \sqrt{1 + \left( \frac{dy}{dx} \right)^2} \, dx \]. In our case, after substituting, it became: \[ L = \int_{0}^{1} \frac{1}{2}(e^x + e^{-x}) \, dx \]. This format is quite manageable since it's essentially an integral of hyperbolic cosine-related terms.Evaluating this, the integral becomes:

• \( \left[ e^x - e^{-x} \right]_{0}^{1} \)

Substituting and simplifying it yields: \[ L = \left( e - \frac{1}{e} \right) \]. Finally, multiplying by \( \frac{1}{2} \) gives the arc length of \[ L = \frac{1}{2} \left( e - \frac{1}{e} \right) \]. This process shows how integral calculus and recognizing specific function properties can efficiently compute such lengths, demonstrating the power and utility of calculus in practical applications.