Problem 129 - Thomas Calculus in SI Units

Question

Find the length of each curve. $$y=\frac{1}{2}\left(e^{x}+e^{-x}\right) \text { from } x=0 \text { to } x=1$$

Step-by-Step Solution

Answer

The length of the curve is \( \frac{1}{2} \left( e - \frac{1}{e} \right) \).

Step 1: Write the Formula for Arc Length

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

Step 2: Calculate the Derivative of the Function

Given the function \( y = \frac{1}{2} \left( e^x + e^{-x} \right) \).First, find the derivative, \( \frac{dy}{dx} \):\[ \frac{dy}{dx} = \frac{1}{2} \left( e^x - e^{-x} \right) \]

Step 3: Plug the Derivative into the Arc Length Formula

Substitute \( \frac{dy}{dx} = \frac{1}{2} \left( e^x - e^{-x} \right) \) into the arc length formula:\[ L = \int_0^1 \sqrt{1 + \left( \frac{1}{2} \left( e^x - e^{-x} \right) \right)^2} \, dx \]

Step 4: Simplify the Expression Under the Square Root

Simplify the expression under the square root:\[ \left( \frac{1}{2} (e^x - e^{-x}) \right)^2 = \frac{1}{4} (e^{2x} - 2 + e^{-2x}) \]Which means the integrand becomes:\[ \sqrt{1 + \frac{1}{4} (e^{2x} - 2 + e^{-2x})} = \sqrt{\frac{1}{2} (e^{2x} + e^{-2x})} \]

Step 5: Substitute the Simplified Expression Back into the Integral

Now substitute the simplified expression back into the integral:\[ L = \int_0^1 \frac{1}{2} (e^x + e^{-x}) \, dx \]

Step 6: Evaluate the Integral

This integral is straightforward to evaluate:\[ \int_0^1 \frac{1}{2} (e^x + e^{-x}) \, dx = \frac{1}{2} \left[ e^x - e^{-x} \right]_0^1 \]Calculate:\[ \frac{1}{2} \left[ (e^1 - e^{-1}) - (e^0 - e^{0}) \right] = \frac{1}{2} (e - \frac{1}{e} - 0) \]

Step 7: Simplify the Final Expression

Finally, simplify to get the length:\[ L = \frac{1}{2} \left( e - \frac{1}{e} \right) \]

Key Concepts

Integral Calculus

Integral calculus is a core topic in mathematics that focuses on finding the total accumulation of quantities. One common use is finding the arc length of a curve. In our problem set, we calculated the arc length of the function \( y = \frac{1}{2}(e^x + e^{-x}) \) from \( x = 0 \) to \( x = 1 \).

• The arc length formula is crucial here: \( L = \int_a^b \sqrt{1 + \left( \frac{dy}{dx} \right)^2}\, dx \).
• This formula accounts for the derivative, \( \frac{dy}{dx} \), which affects the rate of change of the curve.
• Integrating this expression from \( x = 0 \) to \( x = 1 \) gives us the total length of the curve.

Breaking down the mathematical expression into digestible segments makes it easier to handle complex integrals. Integration often requires simplifying expressions under the square root to make the calculations manageable. This comprehension helps in both theoretical and practical aspects of calculus, enhancing problem-solving skills.

Derivative Calculation

Understanding derivatives is essential in calculus as they represent the rate of change of a function. In this problem, we first derive the given function \( y = \frac{1}{2} (e^x + e^{-x}) \).

• To find the derivative \( \frac{dy}{dx} \), apply the derivative rules for exponential functions. The result: \( \frac{dy}{dx} = \frac{1}{2} (e^x - e^{-x}) \).
• Derivatives provide a snapshot of how the curve slopes at any given point, enabling the calculation of arc length.
• Incorporating these derivatives into the arc length formula helps to portray the path a function traces more accurately.

The clarity and precision in calculating derivatives lead to a deeper understanding of how functions behave. Derivatives turn vague, abstract concepts into concrete data points that describe the behavior of a function.

Exponential Functions

Exponential functions, recognizable through their continuous growth or decay patterns, are pivotal in numerous real-world applications. Here, the function \( y = \frac{1}{2} (e^x + e^{-x}) \) incorporates both exponential growth \( e^x \) and decay \( e^{-x} \).

• Exponential functions are defined as \( y = a \cdot e^{kx} \) where \( e \) is a mathematical constant approximately equal to 2.71828.
• The base of the natural logarithm \( e \) aids in the simplification of growth and decay problems.
• In the given function, the terms \( e^x \) and \( e^{-x} \) represent growth and decay, respectively, impacting the curve's behavior and shape.

The combination of these functions results in a hyperbolic cosine function, reflecting a symmetric curve about the y-axis. Recognizing these patterns helps decipher complex equations and understand various scientific phenomena.