Question

In Exercises \(3-14,\) find the arc length of the graph of the function over the indicated interval. $$ y=\ln (\sin x), \quad\left[\frac{\pi}{4}, \frac{3 \pi}{4}\right] $$

Short answer

The arc length of the graph of the function \(y=\ln(\sin(x))\) over the interval \([\frac{\pi}{4}, \frac{3\pi}{4}]\) is \(\ln(3)\).

Step-by-step solution

Calculate the derivative of the function

First, find the derivative of \(y=\ln(\sin(x))\). Using the chain rule, we have \(y'=\frac{1}{\sin(x)}\cdot \cos(x)=\cot(x)\).

Implement the Arc Length formula

Now use the Arc Length formula which is \(L=\int_a^b \sqrt{1+(f'(x))^2} dx\). This involves substituting \(f'(x)\) in the formula with \(\cot(x)\) we obtained in Step 1. Therefore, the arc length is given by \(L=\int_{\frac{\pi}{4}}^{\frac{3\pi}{4}} \sqrt{1+\cot^2(x)} dx\). But, \(\cot^2(x)+1=\csc^2(x)\). So, the integral simplifies to \(L=\int_{\frac{\pi}{4}}^{\frac{3\pi}{4}} |\csc(x)| dx\).

Evaluate the Integral

It is known that the integral of \(|\csc(x)|\) is \(-\ln|\csc(x) + \cot(x)|\). Now, evaluate this over the limits from \(\frac{\pi}{4}\) to \(\frac{3\pi}{4}\). The arc length \(L=-\ln|\csc\left(\frac{3\pi}{4}\right) + \cot\left(\frac{3\pi}{4}\right)| + \ln|\csc\left(\frac{\pi}{4}\right) + \cot\left(\frac{\pi}{4}\right)|\).

Simplify the Result

Simplify the values to get \(L=-\ln|2-1| + \ln|2+1| = -\ln(1) + \ln(3) = \ln(3)\).

Key concepts

Chain Rule

Understanding the chain rule is essential for finding derivatives of composite functions—functions made up of two or more other functions. Think of it like peeling an onion, with each layer representing a function within another function. The chain rule tells us that to differentiate a composite function, we calculate the derivative of the outer function and multiply it by the derivative of the inner function.

For instance, in our exercise, we need to find the derivative of the function \(y = \ln(\sin(x))\). Here, \(\ln(x)\) is the outer function, and \(\sin(x)\) is the inner function. Applying the chain rule, we differentiate \(\ln(x)\), which gives us \(1/x\), and then multiply it by the derivative of \(\sin(x)\), which is \(\cos(x)\). This operation results in \(y' = \cot(x)\), which is the derivative we use in the rest of the problem. The ability to apply the chain rule smoothly can greatly simplify calculus problems, especially when dealing with trigonometric functions and integration.

Integral Calculus

Integral calculus, including arc length calculations, extends beyond basic antiderivatives, engaging with the concept of accumulation. When we calculate arc lengths, we're essentially adding up an infinite number of infinitesimally small line segments along a curve to find its total length.

In our example, to find the arc length of the function \(y = \ln(\sin(x))\) over the interval \([\frac{\pi}{4}, \frac{3\pi}{4}]\), we use the arc length formula \(L = \int_a^b \sqrt{1 + (f'(x))^2} dx\). Integration, in this context, helps compute the total 'distance' covered by the curve. The ability to evaluate integrals accurately is foundational for this kind of calculus problem. It involves recognizing patterns, like knowing that \(\cot^2(x) + 1 = \csc^2(x)\), simplifying complex expressions, and understanding the integral of absolute trigonometric functions, such as \(|\csc(x)|\).

Trigonometric Functions

Trigonometric functions crop up frequently in calculus, especially when dealing with periodic phenomena or geometric problems like arc lengths. They can often complicate integral calculus if one is unfamiliar with their properties and how they interrelate.

In the context of our exercise, recognizing that \(\cot^2(x) + 1 = \csc^2(x)\) is a consequence of the Pythagorean identity and is pivotal in simplifying our integral for arc length. Additionally, familiarity with the integrals of trigonometric functions allows for ready evaluation of the arc length, turning what appears to be a complex integral into a manageable expression. It is this depth of knowledge in trigonometric functions that can transform a daunting calculus problem into a straightforward exercise.