Problem 4 - Calculus: Early Transcendentals

Question

Find the arc length of \(f(x)=\ln (\sin x)\) on the interval \([\pi / 4, \pi / 3]\).

Step-by-Step Solution

Answer

The arc length is \( \ln \left( \frac{1/\sqrt{3}}{\sqrt{2} - 1} \right) \).

Step 1: Understand the Arc Length Formula

The formula for the arc length of a function \(f(x)\) over an interval \([a,b]\) is given by: \[ L = \int_{a}^{b} \sqrt{1 + \left( \frac{df}{dx} \right)^2} \, dx \] We need to find \( \frac{df}{dx} \) for our function \( f(x) = \ln(\sin x)\) to use this formula.

Step 2: Find the Derivative

Calculate the derivative of \( f(x) = \ln(\sin x) \) using the chain rule:\[ \frac{df}{dx} = \frac{1}{\sin x} \cdot \cos x = \cot x. \]

Step 3: Substitute the Derivative

Substitute \( \frac{df}{dx} = \cot x \) into the arc length formula:\[ L = \int_{\pi/4}^{\pi/3} \sqrt{1 + \cot^2 x} \, dx. \]

Step 4: Simplify the Integrand

Recall the trigonometric identity: \[ 1 + \cot^2 x = \csc^2 x. \]Thus, the integrand simplifies to \( \csc x \). The integral becomes:\[ L = \int_{\pi/4}^{\pi/3} \csc x \, dx. \]

Step 5: Integrate the Function

We need to evaluate the integral:\[ \int \csc x \, dx = \ln |\csc x - \cot x| + C. \]So, for the definite integral, we compute:\[ L = \left[ \ln | \csc x - \cot x | \right]_{\pi/4}^{\pi/3}. \]

Step 6: Evaluate the Definite Integral

Calculate the limits:- At \( x = \pi/3 \), \( \csc(\pi/3) = \frac{2}{\sqrt{3}}, \cot(\pi/3) = \frac{1}{\sqrt{3}} \), so: \[ \ln \left( \frac{2}{\sqrt{3}} - \frac{1}{\sqrt{3}} \right) = \ln \left( \frac{1}{\sqrt{3}} \right). \]- At \( x = \pi/4 \), \( \csc(\pi/4) = \sqrt{2}, \cot(\pi/4) = 1 \), so: \[ \ln (\sqrt{2} - 1). \]The arc length is:\[ L = \ln \left( \frac{1}{\sqrt{3}} \right) - \ln (\sqrt{2} - 1). \]

Step 7: Simplify the Result

Simplify the expression using properties of logarithms:\[ L = \ln \left( \frac{1/\sqrt{3}}{\sqrt{2} - 1} \right). \]Evaluate this for simplicity, if possible.

Key Concepts

Derivative

A derivative represents the rate at which a function is changing at any given point. In other words, it's the slope of the function at that specific point. Think of a derivative as a tool that lets you see how steep a curve is. This is crucial in finding the arc length, as it helps us understand how much the curve or line is bending.
Here, we're working with the natural logarithm of the sine function:

• Original function: \( f(x) = \ln(\sin x) \)
• Derivative: \( \frac{df}{dx} = \frac{1}{\sin x} \cdot \cos x = \cot x \).

To find the derivative of \( f(x) \), we employ the chain rule. This rule is vital as it relates derivatives of composed functions. For example, the derivative of \( \ln(\sin x) \) involves differentiating both the outer function (ln) and the inner function (sin) sequentially. This is why derivatives form a fundamental step in calculus, particularly when calculating curve properties like arc length.

Integral Calculus

Integral calculus helps us calculate areas, volumes, central points, and much more. It's essentially about summing up 'parts' to find a whole. In the context of finding the arc length, we use integration to sum up every small segment of the curve over a certain interval, creating a total length.
The arc length formula for the function \( f(x) \) is given by:

• \( L = \int_{a}^{b} \sqrt{1 + \left( \frac{df}{dx} \right)^2} \, dx \)

Once we find the derivative and square it as required by the formula, we integrate this over the specified limits. The arc length on the interval \([\pi/4, \pi/3]\) transforms into a task of evaluating the integral of a function derived in the previous steps. Integration ties this process together by accumulating these tiny sections of the curve we want to measure. This mathematical operation is like stitching pieces of a puzzle together, where each little part contributes to the final answer.

Trigonometric Identities

Trigonometric identities are equations that involve trigonometric functions and are true for every value of the occurring variables. These identities allow us to simplify complex trigonometric expressions. They are particularly useful in calculus to reduce the complexity of an integral, such as those found in arc length problems.
In our example:

• The original integrand \( \sqrt{1 + \cot^2 x} \) simplifies using the identity \( 1 + \cot^2 x = \csc^2 x \)
• This results in the simpler integral \( \int \csc x \, dx \) which can be integrated more swiftly.

Without recognizing and applying these identities, the solution process would be much more challenging and lengthy. By substituting the right identity at the right time, we can transform a problem that seems complicated into one that's much easier to handle. This is crucial in calculus, where simplification often means the difference between solvable and overwhelming equations.