Question

Find the arc length of the function on the given interval. \(f(x)=\frac{1}{2}\left(e^{x}+e^{-x}\right)\) on \([0, \ln 5]\)

Short answer

The arc length is 2.4.

Step-by-step solution

Formula for Arc Length

The arc length of a curve given by a function \( f(x) \) from \( a \) to \( b \) is calculated using the formula: \[L = \int_a^b \sqrt{1 + \left( f'(x) \right)^2} \, dx\]

Find the Derivative of the Function

First, calculate the derivative of the function \( f(x) = \frac{1}{2}(e^x + e^{-x}) \). Using the chain rule, we get: \[f'(x) = \frac{1}{2}(e^x - e^{-x})\]

Simplify the Expression Under the Square Root

Substitute \( f'(x) \) into the arc length formula: \[1 + (f'(x))^2 = 1 + \left(\frac{1}{2}(e^x - e^{-x})\right)^2\]Simplify: \[1 + \frac{1}{4}(e^{2x} - 2 + e^{-2x}) = \frac{4}{4} + \frac{1}{4}(e^{2x} - 2 + e^{-2x}) = \frac{1}{4}(e^{2x} + 2 + e^{-2x})\]The expression becomes: \[\frac{1}{4}(e^x + e^{-x})^2 = f(x)^2\]

Substitute and Integrate

Using the simplified form, our integral becomes: \[L = \int_0^{\ln 5} f(x) \, dx = \int_0^{\ln 5} \frac{1}{2}(e^x + e^{-x}) \, dx\]Calculate the integral: \[L = \left[ \frac{1}{2}(e^x - e^{-x}) \right]_0^{\ln 5}\]Evaluate at the bounds: \[L = \frac{1}{2} \left((5 - \frac{1}{5}) - (1 - 1)\right) = \frac{1}{2}(\frac{24}{5}) = \frac{12}{5} = 2.4\]

Conclusion

Therefore, the arc length of the curve \( f(x) = \frac{1}{2}(e^x + e^{-x}) \) on the interval \([0, \ln 5]\) is 2.4.

Key concepts

Integration

Integration is a fundamental concept in calculus that involves finding the integral of a function. This process can be understood as the opposite of differentiation or as a method to calculate the area under a curve.
When calculating the arc length of a curve, integration is used to add up small segments of a curve from point **a** to point **b**. This total sum gives the exact length of the curve.

• In our exercise, integration helps us calculate the arc length of the function over the specified interval \( [0, \ln 5] \).
• The integration formula takes into account not only the initial function but also its derivative.
• This is because both the shape and slope of the curve are necessary for precise length computation.

Understanding integration as a summation of infinitely small pieces can make finding arc length more intuitive. Each small piece forms a part of the curve, accurately contributing to the entire length.

Derivative

The derivative of a function represents the rate at which the function's value is changing at any point. It shows the slope or steepness of the curve at that particular point.
For arc length, the derivative is crucial because it helps us understand how the curve is behaving.

• In our exercise, the derivative of the function \( f(x) = \frac{1}{2}(e^x + e^{-x}) \) is found using differentiation techniques.
• Specifically, we've used the chain rule to derive \( f'(x) = \frac{1}{2}(e^x - e^{-x}) \).
• This derivative is squared as part of the arc length formula, emphasizing its influence on the length.

Calculating the derivative correctly is vital, as it determines the accuracy of the arc length results. The derivative captures changes in the curve that contribute to its true length.

Chain Rule

The chain rule is an essential method in calculus for finding the derivative of composite functions. A composite function is where one function is nested inside another function.
This rule states that to differentiate a composite function, we must take the derivative of the outer function and multiply it by the derivative of the inner function.

• In our example function, \( f(x) = \frac{1}{2}(e^x + e^{-x}) \), the chain rule is used since it involves exponential functions.
• The function comprises multiple parts: \( e^x \) and \( e^{-x} \), each requiring individual differentiation based on the chain rule.
• Applying the chain rule correctly results in the accurate derivative: \( f'(x) = \frac{1}{2}(e^x - e^{-x}) \).

This careful application of the chain rule ensures each part of the exponential components is handled, providing the foundation for subsequent calculations in the arc length problem.

Exponential Function

Exponential functions, such as \( e^x \) and \( e^{-x} \), are functions where the variable appears in the exponent. These functions are known for their rapid growth or decay.
In the context of our problem, the challenge with exponential functions lies in their different growth rates, making their integration and differentiation slightly different from polynomial functions.

• The function in our exercise, \( f(x) = \frac{1}{2}(e^x + e^{-x}) \), is composed of exponential parts, portraying unique curvature characteristics.
• Exponential functions contribute significantly to the complexity of the arc length formula due to their rapid change rates.
• Understanding the properties of exponentials is critical for integrating them properly over specific intervals, leading to accurate length calculations.

The behavior of exponential functions necessitates precision, especially in calculus, as small changes can significantly impact results. Knowing how exponential functions work aids in understanding their integration and differentiation for accurate solution development.