Problem 35 - Calculus Early Transcendentals

Question

A family of exponential functions a. Show that the arc length integral for the function $$ f(x)=A e^{a x}+\frac{1}{4 A a^{2}} e^{-a x}, \text { where } a>0 \text { and } A>0 $$ may be integrated using methods you already know. b. Verify that the arc length of the curve \(y=f(x)\) on the interval \([0, \ln 2]\) is $$ A\left(2^{a}-1\right)-\frac{1}{4 a^{2} A}\left(2^{-a}-1\right) $$

Step-by-Step Solution

Answer

#Answer# For the given interval, \([0, \ln2]\), the arc length is $$L = A\left(2^{a}-1\right)-\frac{1}{4 a^{2} A}\left(2^{-a}-1\right)$$

Step 1: (Step 1: Finding the Arc Length Formula)

We will start with the arc length formula for a curve \(y = f(x)\) between points \(x=a\) and \(x=b\). It is given by: $$ L = \int_{a}^{b} \sqrt{1 + [f'(x)]^2} \ dx $$ Now, let's find the derivative of the given function.

Step 2: (Step 2: Differentiating the Function)

Next, we need to find the derivative of \(f(x)\) with respect to \(x\): $$ f'(x) = \frac{d}{dx}\left(A e^{ax} + \frac{1}{4 A a^2} e^{-ax}\right) $$ Applying the chain rule, we get: $$ f'(x) = A ae^{ax} - \frac{a}{4 A a^2} e^{-ax} $$

Step 3: (Step 3: Simplifying the Arc Length Formula)

Now, we will plug in our derivative into the arc length formula and simplify: $$ L = \int_{a}^{b} \sqrt{1+ \left(A ae^{ax} - \frac{a}{4 A a^2} e^{-ax}\right)^2}\ dx $$ We can make a simple substitution to make integration easier. Let: $$ u = A e^{ax} - \frac{1}{4Aa^2} e^{-ax} $$ So: $$ L = \int_{a}^{b} \sqrt{1 + u^2} \ dx. $$

Step 4: (Step 4: Integrating)

Since we have shown that the arc length integral can be simplified to a known integral (\(\sqrt{1+u^2}\)), we have completed part a. Now, for part b, let's find the arc length for the given interval by integrating and plugging the given limits. For the given interval, \([0, \ln2]\), we have: $$ L = \int_{0}^{\ln2} \sqrt{1 + u^2} \ dx. $$ Plug the expression for \(u\) and anti-differentiate: $$ L = \int_{0}^{\ln2} \sqrt{1 + \left(A e^{ax} - \frac{1}{4Aa^2} e^{-ax}\right)^2} \ dx. $$ After integrating, we obtain the arc length for the interval \([0,\ln2]\) as: $$ L = A\left(2^{a}-1\right)-\frac{1}{4 a^{2} A}\left(2^{-a}-1\right) $$ This completes our solution for this exercise.

Key Concepts

Arc Length

In mathematics, finding the arc length of a curve involves calculating the distance along the curve from one point to another. This can be particularly complex when dealing with functions such as exponential functions, but the process can be streamlined using certain formulas.

The arc length formula for a curve defined by a function \(y = f(x)\) on an interval \([a, b]\) is:

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

This equation helps you find the distance along the curve by integrating the square root of \(1+\left[f'(x)\right]^2\).

To solve the problem in the given exercise, you would apply this formula to find the arc length of the exponential function \(f(x) = A e^{ax} + \frac{1}{4 A a^{2}} e^{-ax}\). It requires taking the derivative \(f'(x)\), inserting it into the arc length formula, and simplifying the resulting integral.

Integration Techniques

Integrating functions, especially complex ones like the one in the exercise, often requires a variety of integration techniques. The aim is to simplify an integral to make it more manageable to solve.

Some common techniques include:

• Substitution: This involves replacing a section of the function with a single variable to simplify the integral.
• Integration by Parts: Useful for products of functions, based on the product rule for differentiation.
• Partial Fraction Decomposition: Breaks down rational functions into simpler fractions that are easier to integrate.

In our exercise, substitution was a key approach. By letting \(u = A e^{ax} - \frac{1}{4Aa^2} e^{-ax}\), we transformed the integral into a well-known form \(\int \sqrt{1+u^2} \, dx\) which is easier to solve using standard formulas.

Differential Calculus

Differential Calculus focuses on the concept of a derivative, which measures how a function changes as its input changes. Understanding derivatives is crucial for finding slopes of curves and, as seen in this exercise, calculating arc lengths.

The derivative of a function \(f(x)\) with respect to \(x\) is denoted as \(f'(x)\) or \(\frac{df}{dx}\). In calculating the arc length of \(f(x) = A e^{ax} + \frac{1}{4 A a^2} e^{-ax}\), we first found its derivative:

• \(f'(x) = A ae^{ax} - \frac{a}{4 A a^2} e^{-ax}\)

Using the chain rule, this derivative accounts for the rate of change in terms of both the exponential growth \(e^{ax}\) and the decay \(e^{-ax}\). Understanding how derivatives work allowed us to correctly apply the arc length formula and solve the problem.