Problem 4 - Calculus Early Transcendentals

Question

Arc length calculations Find the arc length of the following curves on the given interval by integrating with respect to \(x\). $$y=\frac{1}{2}\left(e^{x}+e^{-x}\right) ;[-\ln 2, \ln 2]$$

Step-by-Step Solution

Answer

To find the arc length of the curve \(y=\frac{1}{2}\left(e^{x}+e^{-x}\right)\) on the interval \([-\ln 2, \ln 2]\), we first determined its derivative, which is \(y'(x) = \frac{1}{2}\left(e^x - e^{-x}\right)\). Then, we plugged the derivative into the arc length formula, resulting in the integral \(L = \int_{-\ln 2}^{\ln 2} \sqrt{\frac{1}{4}\left(e^{2x} + 2 + e^{-2x}\right)} dx\). Unfortunately, this integral cannot be evaluated using elementary functions, so we leave it in this form to represent the arc length of the curve.

Step 1: Find the derivative \(y'(x)\)

To find the derivative of \(y(x)=\frac{1}{2}\left(e^x+e^{-x}\right)\) with respect to \(x\), we differentiate each term in the sum separately: $$y'(x) = \frac{1}{2}\left(\frac{d}{dx}e^x + \frac{d}{dx}e^{-x}\right) = \frac{1}{2}\left(e^x - e^{-x}\right)$$

Step 2: Plug \(y'(x)\) into the arc length formula

Now we will substitute the derivative we found in Step 1 into the arc length formula: $$L = \int_{-\ln 2}^{\ln 2} \sqrt{1+\left(\frac{1}{2}\left(e^x - e^{-x}\right)\right)^2} dx$$

Step 3: Simplify the arc length integrand

To simplify the integrand, we will square the terms inside the square root: $$\left(\frac{1}{2}\left(e^x - e^{-x}\right)\right)^2 = \frac{1}{4}\left(e^{2x} - 2 + e^{-2x}\right)$$ Now we can substitute this back into the integrand and simplify further: $$L = \int_{-\ln 2}^{\ln 2} \sqrt{1+\frac{1}{4}\left(e^{2x} - 2 + e^{-2x}\right)} dx$$ $$L = \int_{-\ln 2}^{\ln 2} \sqrt{\frac{1}{4}\left(e^{2x} + 2 + e^{-2x}\right)} dx$$

Step 4: Evaluate the integral

Unfortunately, this integral cannot be evaluated using elementary functions. We will leave the integral in this form since it best represents the arc length of the curve: $$L = \int_{-\ln 2}^{\ln 2} \sqrt{\frac{1}{4}\left(e^{2x} + 2 + e^{-2x}\right)} dx$$

Key Concepts

Integration with respect to x

Integration is a core concept in calculus, and when we integrate with respect to \(x\), we are essentially finding the accumulated area under a curve as \(x\) changes. In the context of finding arc lengths, we use integration to determine the length of a curve between two points.
The formula for arc length \(L\) of a curve \(y=f(x)\) from \(x=a\) to \(x=b\) is:

• \(L = \int_{a}^{b} \sqrt{1 + (f'(x))^2} \, dx\)

This formula is derived from the Pythagorean theorem, considering an infinitesimally small segment of the curve as a straight line triangle. By summing up these tiny segments from \(x=a\) to \(x=b\), using integration, we get the total arc length.
In the problem, we integrate from \(-\ln 2\) to \(\ln 2\) using the derivative \(y'(x)=\frac{1}{2}(e^x - e^{-x})\), plugged into the arc length formula. This gives us a full representation of the curve's length over the specified interval.

Calculus

Calculus is a branch of mathematics that deals with change and motion. It consists mainly of two components: differentiation and integration. In this exercise, both of these components play a crucial role.
Differentiation is used to find the derivative \(f'(x)\) of a function, which provides the slope of the tangent to the curve at any point \(x\). This slope is essential when calculating the arc length, as it forms part of the arc length integrand formula.
Integration, on the other hand, is used to sum up an infinite number of infinitesimally small quantities, such as the arc length segments mentioned earlier. It allows us to accumulate these small lengths across the interval \([-\ln 2, \ln 2]\) to calculate the total length of the curve. Calculus bridges the gap between algebraic concepts and the geometry of graphs, enabling us to solve complex problems involving curves and rates of change.

Derivative of Exponential Functions

Calculating the derivative of exponential functions is a fundamental skill in calculus, and it's essential for solving the given exercise.
An exponential function is generally of the form \(e^x\), where \(e\) is the mathematical constant approximately equal to 2.718. The unique property of the exponential function \(e^x\) is that its derivative is itself, i.e., \(\frac{d}{dx} e^x = e^x\).

• For the negative exponential, \(\frac{d}{dx} e^{-x} = -e^{-x}\), due to the chain rule of differentiation.

In the exercise, the function is \(y(x) = \frac{1}{2}(e^x + e^{-x})\). The derivative is computed term by term:

• The derivative of the first term \(\frac{1}{2} e^x\) is \(\frac{1}{2} e^x\).
• The derivative of the second term \(\frac{1}{2} e^{-x}\) is \(-\frac{1}{2} e^{-x}\).

Thus, combining these gives \(y'(x) = \frac{1}{2}(e^x - e^{-x})\), which is then used to compute the arc length. Understanding these derivatives helps in manipulating exponential functions in calculus and solving related problems efficiently.