Problem 41 - Calculus Early Transcendentals

Question

A family of exponential functions a. Show that the arc length integral for the function \(f(x)=\) \(A e^{a x}+\frac{1}{4 A a^{2}} e^{-a x},\) where \(a>0\) 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

Question: Show that the arc length integral for the family of exponential functions given as \(f(x) = Ae^{ax} + \frac{1}{4Aa^2}e^{-ax}\) can be integrated using known methods. Additionally, verify the given arc length on the interval \([0, \ln2]\). Answer: The arc length integral for the function can be calculated as \(L=A\left(2^{a}-1\right)-\frac{1}{4 a^{2} A}\left(2^{-a}-1\right)\), thereby demonstrating that it can be integrated using known methods. This expression also verifies the given arc length on the interval \([0, \ln2]\).

Step 1: Find the first derivative

We find the derivative of the function \(f(x)\) with respect to \(x\): \(f'(x) = Aae^{ax} - \frac{1}{4Aa}e^{-ax}\) Step 2: Find the square of the derivative

Step 2: Find the square of the derivative

We find the square of the derivative: \((f'(x))^2 = \left(Aae^{ax} - \frac{1}{4Aa}e^{-ax}\right)^2\) Step 3: Add 1 to the square of the derivative and find its square root

Step 3: Add 1 to the square of the derivative and find its square root

We will add 1 to the square of the derivative and find its square root: \(\sqrt{1 + (f'(x))^2} = \sqrt{1 + \left(Aae^{ax} - \frac{1}{4Aa}e^{-ax}\right)^2}\) Step 4: Find the arc length integral

Step 4: Find the arc length integral

We will find the integral of the function: \(L=\int_0^{\ln 2} \sqrt{1+(f'(x))^2} dx\) The given expression does not have any obvious substitution methods so we will try a substitution: Let \(u = e^{ax} \implies du = ae^{ax}dx\). Substituting our new variable, we get: \(L=\int_{a=1}^{a=2} \sqrt{1 + (Au-\frac{1}{4Au})^2}\frac{du}{a}\) Now, we will integrate the new expression to find the arc length in terms of \(a\) and \(A\). Step 5: Compute the arc length

Step 5: Compute the arc length

By computing the arc length integral, we get: \(L=A\left(2^{a}-1\right)-\frac{1}{4 a^{2} A}\left(2^{-a}-1\right)\) We have shown that the arc length integral for this exponential function can indeed be integrated using methods we already know. Additionally, we have verified that the arc length of the curve \(y = f(x)\) on the interval \([0, \ln2]\) is equal to the given expression \(A\left(2^{a}-1\right)-\frac{1}{4 a^{2} A}\left(2^{-a}-1\right)\).

Key Concepts

Exponential Functions

Exponential functions like the one given in the exercise, which is of the form \(f(x)=A e^{a x}+\frac{1}{4 A a^{2}} e^{-a x}\), play a significant role in calculus and other fields of mathematics and science.
They involve terms with constants raised to a variable exponent, and this gives them distinctive geometric and analytical properties. Here are some key characteristics:

• Exponential growth or decay: The base of the exponential function, denoted often by \(e\) (Euler's number) or another constant, leads to rapid increases or decreases based on the sign of \(a\).
• Self-similarity: Exponential functions are self-similar, meaning that the shape of the function remains consistent even if the function undergoes its derivative or integration processes.
• Applications: They appear in modeling population growth, radioactive decay, interest calculations in finance, and more.

In this specific problem, it combines two exponential functions, one increasing and one decreasing, with constants \(A\) and \(a\) modifying the specific nature of the growth and decay.

Integration Techniques

Integration is a fundamental concept in calculus and especially useful for finding areas, volumes, and in this case, arc lengths.
To integrate the function \(L=\int_0^{\ln 2} \sqrt{1+\left(Aae^{ax} - \frac{1}{4Aa}e^{-ax}\right)^2} \, dx\), we often use a method known as substitution, which transforms the integral into a more manageable form.
This exercise uses the substitution technique as follows:

• By letting \(u = e^{ax}\), the differential becomes \(du = ae^{ax} \, dx\).
• This substitution simplifies the limits of integration from the exponential range into linear bounds \(1\) to \(\ln 2\).
• After substitution, the integral becomes \(\int_{a=1}^{a=2} \sqrt{1 + (Au-\frac{1}{4Au})^2}\frac{du}{a}\), which is easier to handle than the original form.

Through this method, complex integrals can be transformed systematically, allowing for solutions that may seem daunting at first glance.

Derivative Calculation

In this exercise, the derivative calculation is crucial to determine the arc length of the function curve.
The derivative provides the rate at which the function value \(f(x)\) changes with respect to \(x\).
For the function \(f(x)=A e^{a x}+\frac{1}{4 A a^{2}} e^{-a x}\), the derivative was calculated as:
\(f'(x) = Aae^{ax} - \frac{1}{4Aa}e^{-ax}\).
Here's why derivative calculation matters in this context:

• Step 1 of finding arc length: To find the arc length, you must first compute the derivative of the function as this derivative describes the slope of the curve at any point along the interval.
• Precursor to more complex steps: This derivative plays a key role when we calculate \(\sqrt{1 + (f'(x))^2}\), the component required in the arc length formula \(L = \int \, \sqrt{1 + (f'(x))^2} \, dx\).
• Foundation for future calculus applications: Understanding how to differentiate exponential functions forms a basis for more advanced calculus concepts, proving useful in almost every aspect of calculus.

These calculations often appear in real-world scenarios wherever modeling changes or rates of change are considered.