Question

In Exercises \(3-14,\) find the arc length of the graph of the function over the indicated interval. $$ y=\frac{1}{2}\left(e^{x}+e^{-x}\right), \quad[0,2] $$

Short answer

The arc length of the graph of the function \(\frac{1}{2}\left(e^{x}+e^{-x}\right)\) over the interval [0, 2] is \(\frac{1}{2}\left(e^{2} - e^{-2}\right).\).

Step-by-step solution

Calculate the derivative

The first step requires calculating the derivative of the given function. In our case, we differentiate \(y=\frac{1}{2}\left(e^{x}+e^{-x}\right)\). This gives us \(y'= \frac{1}{2}\left(e^{x}-e^{-x}\right)\). The power rule for differentiation is applied here.

Square the derivative

Next, we square the calculated derivative, i.e., \((y')^{2}= \left(\frac{1}{2}\left(e^{x}-e^{-x}\right)\right)^{2} = \frac{1}{4}\left(e^{2x}+ e^{-2x}- 2\right). These manipulations are done according to the rules of powers and exponents.

Find the square root of the sum of 1 and the squared derivative

We then find the square root of \(1 + (y')^{2}\), that is, \(\sqrt{1 + \frac{1}{4}\left(e^{2x}+ e^{-2x}- 2\right)} = \frac{1}{2}\left(e^{x}+e^{-x}\right) = y. Interesting to notice is that the function under the square root simplifies to the original function.

Compute the integral

Finally, we integrate the result from the previous step over the given interval [0, 2]. That is, \(\int_{0}^{2} y \, dx = \int_{0}^{2} \frac{1}{2}\left(e^{x}+e^{-x}\right) \, dx\). Using the rules of integration with respect to each term individually, we get \(\frac{1}{2}\left[e^{x}-e^{-x}\right]_{0}^{2} = \frac{1}{2}\left(e^{2} - e^{-2}\right) - \frac{1}{2}\left(1 - 1\right) = \frac{1}{2}\left(e^{2} - e^{-2}\right).\)

Simplify the result

We simplify the equation to find the arc length, which leads us to \(\frac{1}{2}\left(e^{2} - e^{-2}\right).\)

Key concepts

Differentiation

Differentiation is a fundamental concept in calculus that concerns the process of determining the rate at which a function is changing at any given point. In simpler terms, it measures how much the output of a function changes when its input changes ever so slightly. This process is called finding the derivative of the function.

Let's take a closer look at our exercise. The function given is \(y = \frac{1}{2}(e^x + e^{-x})\). To find the arc length, we first need to differentiate this function. By applying the differentiation rules for exponential functions, the derivative of \(e^x\) is \(e^x\), and the derivative of \(e^{-x}\) is \(-e^{-x}\). The derivative of our function is hence \(y' = \frac{1}{2}(e^x - e^{-x})\). Knowing how to differentiate correctly is crucial for the next steps in solving arc length calculus problems.

Integration

Integration, the inverse process of differentiation, is another core calculus tool. It involves finding a function when its derivative is known, and it's also used to calculate areas under curves and, as in this case, the arc length of curves.

After differentiation and simplifying expressions to find the formula for the arc length, we need to integrate the function over the given interval to actually calculate it. In our exercise, we integrate the simplified expression of the derivative over the interval [0,2]. The integration of \(e^x\) results in \(e^x\), and for \(e^{-x}\), it results in \(-e^{-x}\). The integration process requires a careful application of rules and properties of integrals to ensure the correct arc length is found.

Exponential Functions

Exponential functions have the form \(f(x) = a^x\), where the base \(a\) is a constant, and the exponent \(x\) is a variable. These functions are known for their unique properties, such as growth and decay, and they are particularly important in modeling real-world situations like compounding interest, population growth, and radioactive decay.

In our arc length problem, we work with the exponential functions \(e^x\) and \(e^{-x}\), which are commonly seen in calculus due to the natural exponential function's constant appearance in continuous growth and decay models. Recognizing and being comfortable with manipulating these functions is crucial for solving calculus problems that involve them, such as finding the arc length of their graphs.

Calculus Problems

Solving calculus problems often involves a series of steps that require a combination of differentiation and integration techniques, along with algebraic manipulations. Knowing and understanding these concepts, as well as how they interrelate, is vital for approaching these problems successfully.

In the example given, finding the arc length involves differentiating the function, squaring the derivative, adding one, taking the square root, and then integrating the result over an interval. This problem typifies the multi-faceted approach needed for calculus problems and highlights the importance of a step-by-step method in solving them. By breaking down the problem into smaller parts and tackling each step carefully, students can arrive at the correct solution more easily.