Problem 18 - Calculus: Early Transcendentals

Question

Find the arc length of the function on the given interval. $$ f(x)=\ln (\sin x) \text { on }[\pi / 6, \pi / 2] $$

Step-by-Step Solution

Answer

The arc length is \( \ln(2+\sqrt{3}) \).

Step 1: Understand the Formula for Arc Length

The arc length of a function over an interval \([a, b]\) is given by the formula: \[ L = \int_a^b \sqrt{1 + \left( f'(x) \right)^2} \, dx \] where \( f'(x) \) is the derivative of the function \( f(x) \). In this exercise, \( f(x) = \ln(\sin(x)) \) and the interval is \([\pi/6, \pi/2]\).

Step 2: Find the Derivative of the Function

To find the arc length, first compute the derivative of \( f(x) = \ln(\sin(x)) \). Using the chain rule: \[ f'(x) = \frac{1}{\sin(x)} \cdot \cos(x) = \cot(x) \]

Step 3: Substitute the Derivative into the Arc Length Formula

Next, substitute \( f'(x) = \cot(x) \) into the arc length formula: \[ L = \int_{\pi/6}^{\pi/2} \sqrt{1 + \cot^2(x)} \, dx \] From the trigonometric identity, we know \( 1 + \cot^2(x) = \csc^2(x) \). This simplifies the integrand to: \( \csc(x) \).

Step 4: Integrate to Find the Arc Length

Now, evaluate the integral: \[ L = \int_{\pi/6}^{\pi/2} \csc(x) \, dx \] The antiderivative of \( \csc(x) \) is \( -\ln|\csc(x) + \cot(x)| \). Thus, the definite integral is: \[ L = \left[ -\ln|\csc(x) + \cot(x)| \right]_{\pi/6}^{\pi/2} \]

Step 5: Evaluate the Definite Integral

Compute the definite integral: - At \( x = \pi/2 \): \( -\ln|\csc(\pi/2) + \cot(\pi/2)| = -\ln|1 + 0| = 0 \) - At \( x = \pi/6 \): \( -\ln|\csc(\pi/6) + \cot(\pi/6)| = -\ln|2 + \sqrt{3}| \)So, \[ L = 0 - \left(-\ln|2 + \sqrt{3}|\right) = \ln(2 + \sqrt{3}) \]

Step 6: Conclusion

The arc length of the curve \( f(x) = \ln(\sin(x)) \) from \( \pi/6 \) to \( \pi/2 \) is \( \ln(2 + \sqrt{3}) \).

Key Concepts

Derivative

When dealing with functions like \( f(x) = \ln(\sin x) \), derivatives are crucial, especially for computing arc lengths. The derivative of a function gives us a way to understand how the function changes at any given point. For \( f(x) = \ln(\sin x) \), finding the derivative requires the use of the chain rule.

• The chain rule helps differentiate composed functions, like \( \ln(\sin x) \).
• To apply it: differentiate the outer function \( \ln u \) to get \( \frac{1}{u} \), then multiply by the derivative of the inner function \( \sin x \), which is \( \cos x \).

This process gives us \( f'(x) = \frac{1}{\sin x} \cdot \cos x = \cot x \), indicating that the rate of change of the logarithm of sine is essentially the cotangent of \( x \). Understanding this derivative is key when integrating to find the arc length.

Trigonometric Identity

Trigonometric identities are vital tools in simplifying and solving problems involving trigonometric functions. In the context of finding arc lengths, they help simplify expressions before integration. A significant identity used here is:

• \( 1 + \cot^2(x) = \csc^2(x) \)

This identity transforms the expression under the square root of the arc length formula. The original form \( \sqrt{1+\cot^2(x)} \) becomes \( \csc(x) \). This simplification streamlines integration. It's crucial because it changes a potentially complex integral into a form where the integral of \( \csc(x) \) has a known antiderivative, making the whole process more manageable. Utilizing such identities turns challenging problems into ones that are easier to handle, providing a clearer path to the solution.

Definite Integral

The definite integral is a concept used to find the total accumulation of a quantity, such as the arc length over a specific interval. In this problem, the arc length from \( \pi/6 \) to \( \pi/2 \) is calculated using a definite integral.

• We start with the integral \( \int_{\pi/6}^{\pi/2} \csc(x) \, dx \), simplified using trigonometric identities.
• The antiderivative of \( \csc(x) \) is \( -\ln|\csc(x) + \cot(x)| \).

Evaluating this from \( \pi/6 \) to \( \pi/2 \) involves substituting these bounds into the antiderivative.The process includes calculating the value at each limit and subtracting:

• At \( x = \pi/2 \), the function simplifies to 0, contributing no arc length.
• At \( x = \pi/6 \), it calculates \( -\ln|2 + \sqrt{3}| \), a natural logarithm involving the cosecant and cotangent values at \( \pi/6 \).

This final computation gives the arc length as \( \ln(2 + \sqrt{3}) \), illustrating how the definite integral encapsulates the entire arc of the function over the given interval.