Problem 132 - Thomas Calculus in SI Units

Question

Find the length of each curve. $$y=\ln (\csc x) \text { from } x=\pi / 6 \text { to } x=\pi / 4$$

Step-by-Step Solution

Answer

The length of the curve is approximately 0.34657 units.

Step 1: Understand the problem

We're asked to find the length of the curve defined by the function \( y = \ln(\csc x) \) from \( x = \frac{\pi}{6} \) to \( x = \frac{\pi}{4} \). For this, we'll utilize the formula for arc length of a function.

Step 2: Arc length formula

The formula for the length \( L \) of a curve \( y = f(x) \) from \( x = a \) to \( x = b \) is given by: \[ L = \int_{a}^{b} \sqrt{1 + \left( \frac{dy}{dx} \right)^2} \, dx \] Here, \( a = \frac{\pi}{6} \) and \( b = \frac{\pi}{4} \).

Step 3: Differentiate the function

First, we need to find \( \frac{dy}{dx} \) of the function \( y = \ln(\csc x) \). Recall that \( \csc x = \frac{1}{\sin x} \), thus \( y = \ln(\frac{1}{\sin x}) = -\ln(\sin x) \). Differentiating: \( \frac{dy}{dx} = -\frac{1}{\sin x} \cdot \cos x = -\cot x \)

Step 4: Compute the expression under the square root

Substitute \( \frac{dy}{dx} = -\cot x \) into the arc length formula. \[ 1 + \left( \frac{dy}{dx} \right)^2 = 1 + (-\cot x)^2 = 1 + \cot^2 x \] Recall the trigonometric identity: \( 1 + \cot^2 x = \csc^2 x \). Thus, \( 1 + \left( \frac{dy}{dx} \right)^2 = \csc^2 x \).

Step 5: Set up the integral

Substitute into the arc length formula: \[ L = \int_{\pi/6}^{\pi/4} \sqrt{\csc^2 x} \, dx = \int_{\pi/6}^{\pi/4} \csc x \, dx \].

Step 6: Evaluate the integral

We need to evaluate the integral \( \int \csc x \, dx \). The antiderivative of \( \csc x \) is \( -\ln| \csc x + \cot x| \). Thus: \[ L = \left[ -\ln| \csc x + \cot x| \right]_{\pi/6}^{\pi/4} \].

Step 7: Compute the definite integral

Substituting the limits into the antiderivative: \[-\ln| \csc \frac{\pi}{4} + \cot \frac{\pi}{4}| - \left(-\ln| \csc \frac{\pi}{6} + \cot \frac{\pi}{6}|\right) \] At \( \frac{\pi}{4} \), \( \csc x = \sqrt{2} \) and \( \cot x = 1 \). At \( \frac{\pi}{6} \), \( \csc x = 2 \) and \( \cot x = \sqrt{3} \). Therefore, \[ L = -\ln(\sqrt{2} + 1) + \ln(2 + \sqrt{3}) \] \[ L = \ln\left(\frac{2 + \sqrt{3}}{\sqrt{2} + 1}\right) \]

Step 8: Simplify the expression

The expression simplifies to an approximate numerical value using a calculator: \[ L \approx 0.34657 \].

Key Concepts

Differentiation

Differentiation is a fundamental concept in calculus used to determine the rate at which a function is changing at any given point. In this problem, we need the derivative of the function \( y = \ln(\csc x) \). To find it, we recall that \( \csc x = \frac{1}{\sin x} \), which can be rewritten as \( y = \ln(\frac{1}{\sin x}) = -\ln(\sin x) \). Differentiating \( -\ln(\sin x) \) with respect to \( x \) involves two steps: applying the chain rule and the derivative of the natural logarithm.

• The chain rule is used because ln(sin x) is a composition of functions.
• The derivative of \( \ln u \) with respect to \( u \) is \( \frac{1}{u} \).

By applying these, we find \( \frac{dy}{dx} = -\frac{1}{\sin x} \cdot \cos x = -\cot x \). Differentiating functions accurately is crucial as it forms the basis for more complex calculus concepts like the arc length formula here.

Trigonometric Identities

Trigonometric identities help simplify expressions and solve equations involving trigonometric functions. These identities are equations that are universally true for trigonometric functions. In this exercise, calculating arc length involves using the identity \( 1 + \cot^2 x = \csc^2 x \). This identity allows us to simplify the expression \( 1 + \left( \frac{dy}{dx} \right)^2 \) into \( \csc^2 x \), which is much easier to integrate.

• Recognizing and using trigonometric identities can simplify the integration process and make calculations manageable.
• \( \csc x \) and \( \cot x \) are reciprocals of \( \sin x \) and \( \tan x \) respectively, offering alternative ways to handle derivatives and integrals.

Trigonometric identities play a vital role in calculus problems involving curves, making them an essential part of your calculus toolkit.

Integral Calculus

Integral calculus is concerned with accumulation, areas under curves, and finding total quantities from rates of change. In the context of this problem, we use integral calculus to find the arc length of a curve. Once we know \( \csc x \) is the power under the square root, we evaluate the definite integral from \( x = \frac{\pi}{6} \) to \( x = \frac{\pi}{4} \), which corresponds to \( \int_{\pi/6}^{\pi/4} \csc x \, dx \).

• The antiderivative of \( \csc x \) is \( -\ln| \csc x + \cot x| \), a result you'll use frequently in similar problems.
• Computing definite integrals involves substituting limits into the antiderivative and simplifying the result.

Integral calculus allows us to handle various complex phenomena and provides solutions to problems that involve adding up an infinite number of infinitesimal quantities, like arc lengths in this case.