Problem 142 - Thomas Calculus

Question

find the length of each curve. \begin{equation}y=\ln (\csc x) \text { from } x=\pi / 6 \text { to } x=\pi / 4\end{equation}

Step-by-Step Solution

Answer

The length of the curve is \( L = \ln(\sqrt{2}) \).

Step 1: Understand the Formula

The formula for the length of a curve for a function \( y = f(x) \) from \( x = a \) to \( x = b \) is given by \[L = \int_a^b \sqrt{1 + \left( \frac{dy}{dx} \right)^2} \, dx.\] We need to use this formula to find the length of the curve \( y = \ln(\csc x) \) from \( x = \pi/6 \) to \( x = \pi/4 \).

Step 2: Compute the Derivative \( \frac{dy}{dx} \)

First, we need to find the derivative \( \frac{dy}{dx} \) of the function \( y = \ln(\csc x) \). We know that the derivative of \( \ln(\csc x) \) is given by \[\frac{dy}{dx} = -\cot x.\] Thus, \( \frac{dy}{dx} = -\cot x \).

Step 3: Compute \( \left( \frac{dy}{dx} \right)^2 \)

We compute the square of the derivative: \[\left( \frac{dy}{dx} \right)^2 = (-\cot x)^2 = \cot^2 x.\]

Step 4: Set Up the Integral for Length

Now, substitute \( \frac{dy}{dx} \) into the formula for curve length:\[L = \int_{\pi/6}^{\pi/4} \sqrt{1 + \cot^2 x} \, dx.\]Recall that \( 1 + \cot^2 x = \csc^2 x \), so the integral becomes:\[L = \int_{\pi/6}^{\pi/4} \csc x \, dx.\]

Step 5: Evaluate the Integral

The integral of \( \csc x \) is:\[\int \csc x \, dx = -\ln |\csc x + \cot x| + C.\]Thus, \[ L = \left[-\ln|\csc x + \cot x|\right]_\frac{\pi}{6}^\frac{\pi}{4}. \]

Step 6: Calculate the Definite Integral

Substitute the limits into the integrated function:\[L = -\ln|\csc(\pi/4) + \cot(\pi/4)| + \ln|\csc(\pi/6) + \cot(\pi/6)|.\]We know \( \csc(\pi/4) = \sqrt{2}, \cot(\pi/4) = 1, \csc(\pi/6) = 2, \) and \( \cot(\pi/6) = \sqrt{3} \). Therefore,\[L = -\ln|\sqrt{2} + 1| + \ln|2 + \sqrt{3}|.\]Finally:\[L = \ln\left( \frac{2 + \sqrt{3}}{\sqrt{2} + 1} \right).\]

Step 7: Simplify the Expression

Simplifying the expression further, we find that the length of the curve is\[ L = \ln\left( \sqrt{2} \right).\]

Key Concepts

Differentiation

Differentiation is a fundamental concept in calculus that deals with finding the rate of change of a function. In our problem, we're interested in the function

• \(y = \ln(\csc x)\).

To find the arc length, we first need to calculate the derivative \( \frac{dy}{dx} \). The derivative of a logarithmic function requires applying chain rule and knowledge of trigonometric derivatives. For example:

• The derivative of \(\ln f(x)\) is \(\frac{1}{f(x)} \cdot f'(x)\).
• Here, \(f(x) = \csc x\), with a derivative \(-\csc x \cot x\).

Thus, \( \frac{dy}{dx} = -\cot x\).
Differentiation helps us break down complex curves into their simplest components by revealing how the rate of change behaves at every point along the curve. Understanding this makes it easier to deal with various calculus operations like finding arc lengths.

Integral Calculus

Integral calculus is the branch of calculus focused on finding the total accumulation of quantities and the area under curves. When it comes to arc length, integral calculus helps compute the total length of a curve over a specific interval.
To do this, we use the formula for the length of the curve:

• \[L = \int_a^b \sqrt{1 + \left( \frac{dy}{dx} \right)^2} \, dx\].

This integral is evaluated over the interval from \(x = \pi/6\) to \(x = \pi/4\).
Before integration, we substitute the previously found derivative squared, which gives us:

• \[\left( \frac{dy}{dx} \right)^2 = \cot^2 x\].

Using a trigonometric identity, the integral simplifies as \( \sqrt{1 + \cot^2 x} = \csc x\). This lets us evaluate the integral, leading to the solution.
Integral calculus ties these steps together, providing a method to measure curves in real-world contexts and making the abstract concept of arc length applicable.

Trigonometric Identities

Trigonometric identities are equations involving trigonometric functions that hold true for all values of the variable. They are crucial when solving calculus problems that involve trigonometric functions. In our exercise, several identities were used:

• The key identity \(1 + \cot^2 x = \csc^2 x\) helped transform the integrand, simplifying the arc length formula.
• Additionally, values like \(\csc(\pi/4)\) and \(\cot(\pi/4)\) needed their respective identities to evaluate the integral.

These identities enabled us to reduce complex expressions into simpler forms, making the integration step more manageable.
By understanding how to apply these identities effectively, we can often find shortcuts that simplify solving integrals and derivatives involving trigonometric functions. This skill is particularly valuable in calculus, where such functions are frequently encountered.