Question

Find the length of the curve \(y=\ln (\sin x), \frac{\pi}{3} \leq x \leq \frac{\pi}{2}\).

Short answer

The length of the curve is \(\frac{1}{2}\ln 3\).

Step-by-step solution

Understanding the Problem

To find the length of the curve given by the function \(y = \ln(\sin x)\) from \(x = \frac{\pi}{3}\) to \(x = \frac{\pi}{2}\), we will use the formula for the arc length of a function \(y = f(x)\), which is \(\int_{a}^{b} \sqrt{1 + (f'(x))^2} \, dx\).

Finding the Derivative

First, compute the derivative of \(y = \ln(\sin x)\). Using the chain rule and the derivatives \(\frac{d}{dx}[\ln u] = \frac{1}{u}\cdot \frac{du}{dx}\) and \(\frac{d}{dx}[\sin x] = \cos x\), we have:\[y' = \frac{d}{dx}[\ln(\sin x)] = \frac{1}{\sin x} \cdot \cos x = \cot x.\]

Applying the Arc Length Formula

Substitute \(y' = \cot x\) into the arc length formula. The integral becomes:\[\int_{\pi/3}^{\pi/2} \sqrt{1 + (\cot x)^2} \, dx.\]Using the identity \(1 + \cot^2 x = \csc^2 x\), the integral simplifies to:\[\int_{\pi/3}^{\pi/2} \csc x \, dx.\]

Evaluating the Integral

To evaluate \(\int \csc x \, dx\), use the formula:\[\int \csc x \, dx = -\ln|\csc x + \cot x| + C.\]Thus, evaluating from \(\pi/3\) to \(\pi/2\), we get:\[\left[ -\ln|\csc x + \cot x| \right]_{\pi/3}^{\pi/2}.\]

Calculating the Result

Substitute \(x = \frac{\pi}{2}\) and \(x = \frac{\pi}{3}\) into the evaluated expression:1. For \(x = \frac{\pi}{2}\): \(\csc(\frac{\pi}{2}) = 1\), \(\cot(\frac{\pi}{2}) = 0\), so: \[-\ln |1 + 0| = 0.\]2. For \(x = \frac{\pi}{3}\): \(\csc(\frac{\pi}{3}) = \frac{2}{\sqrt{3}}\), \(\cot(\frac{\pi}{3}) = \frac{1}{\sqrt{3}}\), so: \[-\ln \left| \frac{2}{\sqrt{3}} + \frac{1}{\sqrt{3}} \right| = -\ln \left( \frac{3}{\sqrt{3}} \right) = -\ln \sqrt{3}.\]Thus, the length of the curve is:\[0 - [-\ln \sqrt{3}] = \ln \sqrt{3} = \frac{1}{2}\ln 3.\]

Concluding the Solution

Therefore, the length of the curve \(y = \ln(\sin x)\) from \(x = \frac{\pi}{3}\) to \(x = \frac{\pi}{2}\) is:\[\frac{1}{2}\ln 3.\]

Key concepts

Logarithmic Functions

Logarithmic functions are essential mathematical tools often appearing in calculus problems. The function we deal with here, \(y = \ln(\sin x)\), illustrates how logarithms can be combined with trigonometric expressions. A logarithm essentially asks, "To what exponent must a certain base be raised to produce a given number?" In the natural logarithm \(\ln\), the base is \(e\), Euler's number, approximately equal to 2.718. This natural logarithm is particularly useful in calculus for its derivative properties.

In this particular problem, using the natural logarithm allows us to explore the relationship between exponential growth and sinusoidal functions. Understanding this combination is crucial because this pattern appears in many real-world phenomena, like sound waves and alternating currents.

- Logarithms transform multiplicative processes into additive ones, which can simplify the problem-solving process.- They also have special qualities under differentiation, making them a practical choice for calculus.

Trigonometric Identities

Trigonometric identities are pivotal in simplifying calculus problems, especially those involving trigonometric functions. In our exercise, we used the identity \(1 + \cot^2 x = \csc^2 x\). This identity allowed us to simplify the expression under the square root in the arc length integral.

This simplification was crucial in reducing complexity. When solving calculus problems, knowing these identities can save valuable time and effort. Learning these identities can help you swiftly transform and solve equations by simplifying expressions.

• \(\cot x\) is the trigonometric function representing the reciprocal of the tangent: \(\cot x = \frac{1}{\tan x} = \frac{\cos x}{\sin x}\).
• \(\csc x\), or cosecant, is the reciprocal of the sine function: \(\csc x = \frac{1}{\sin x}\).

Understanding these basic identities and how to apply them in calculus can be immensely beneficial, especially as trigonometric functions frequently appear in real-world applications.

Integration Techniques

Integration techniques help us calculate areas, arc lengths, and more. In this problem, finding the arc length involved evaluating the integral \(\int_{\pi/3}^{\pi/2} \csc x \, dx\). This requires using integration formulas and techniques that deal specifically with complex trigonometric functions.

One powerful technique is recognizing how integration formulas, like \(\int \csc x \, dx = -\ln|\csc x + \cot x| + C\), can be applied. By knowing these formulas, you can directly tackle challenging integrals without needing a step-by-step simplification every time.

Some concepts to remember:- Integration often involves finding an antiderivative, which is the reverse process of differentiation.- For more complicated expressions, substitution or parts may be used, but recognizing standard forms can provide a shortcut.

In practice, integration opens the door to solving a host of physical problems, from calculating distances and areas to predicting behavior over time in physics and engineering.

Calculus Problem Solving

Solving calculus problems often means piecing together multiple mathematical concepts. In this exercise, finding the arc length of \(y = \ln(\sin x)\) involved differentiation, applying trigonometric identities, and evaluating integrals.

Effective problem-solving in calculus begins with a clear understanding of the problem and mapping a path to the solution. Start by determining what needs to be calculated; here, it was the curve length. Then, break it down into smaller parts, as seen in the solution steps:- Compute derivatives to understand changes in your function.- Simplify your equations using known identities.- Integrate to find accumulated or total values, like lengths or areas.

As you gain more experience, you'll recognize patterns and strategies that will guide you through various types of problems. Remember, practice and familiarity with fundamental concepts and techniques are key to becoming proficient in calculus.