Problem 10 - Calculus Single Variable

Question

Calculate the arc length \(L\) of the graph of the given function over the given interval. $$ f(x)=\ln (\sin (x)) \quad I=[\pi / 4, \pi / 2] $$

Step-by-Step Solution

Answer

The arc length is \( \ln(\sqrt{2} + 1). \)

Step 1: Recall the Arc Length Formula

The formula to compute the arc length \( L \) of a function \( y = f(x) \) from \( x = a \) to \( x = b \) is given by \[ L = \int_{a}^{b} \sqrt{1 + \left( f'(x) \right)^2} \, dx \]. In this exercise, the function is \( f(x) = \ln(\sin(x)) \), and the interval \( I = [\frac{\pi}{4}, \frac{\pi}{2}] \).

Step 2: Find the Derivative of the Function

Differentiate the function \( f(x) = \ln(\sin(x)) \). Using the chain rule, \( f'(x) = \frac{d}{dx} \ln(\sin(x)) = \frac{1}{\sin(x)} \cdot \cos(x) = \cot(x) \).

Step 3: Set Up the Arc Length Integral

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

Step 4: Simplify the Integral Using Trigonometric Identity

Use the identity \( 1 + \cot^2(x) = \csc^2(x) \). Thus the integral becomes \[ L = \int_{\pi/4}^{\pi/2} \csc(x) \, dx \].

Step 5: Evaluate the Integral

The integral \( \int \csc(x) \, dx \) is \( -\ln| \csc(x) + \cot(x) | + C \). Evaluate this from \( \pi/4 \) to \( \pi/2 \).

Step 6: Compute Final Answer

Calculate \( -\ln| \csc(\pi/2) + \cot(\pi/2) | + \ln| \csc(\pi/4) + \cot(\pi/4) | \). Plug in \( \csc(\pi/2) = 1 \) and \( \cot(\pi/2) = 0 \), and \( \csc(\pi/4) = \sqrt{2} \), \( \cot(\pi/4) = 1 \). This gives \[ L = -\ln|1| + \ln| \sqrt{2} + 1 | = \ln( \sqrt{2} + 1 ). \]

Key Concepts

Integral Calculus

Integral Calculus is a fundamental part of calculus that deals with the concept of integration. Integration is essentially the reverse process of differentiation, every integral is linked to a derivative.
It focuses on finding the total accumulation of quantities.In the context of our exercise, the arc length is an application of integral calculus. The formula for arc length involves integrating over a range. This integration accounts for every little segment of the curve summed to find the total length.
The integral used is

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

where \( f'(x) \) is the derivative of the function. The square root and squaring of \( f'(x) \) ensure that the contributions from each tiny piece of the curve are correctly combined.
This method allows us to measure curves that aren't straight, providing insights and calculations in numerous scientific and engineering applications.

Trigonometric Identities

Trigonometric identities are formulas involving trigonometric functions that are true for every value of the involved variables. These identities are used extensively to simplify expressions and solve trigonometric equations.
In our solution, the identity

• \( 1 + \cot^2(x) = \csc^2(x) \)

is vital. It transforms the expression under the square root into a simpler, recognizable form that can be integrated.Understanding these identities allows us to convert complex trigonometric integrals into simpler forms, making computations more straightforward. Trigonometric identities are particularly powerful since they reduce the complication by substituting known equivalents.
This strategic use of identities provides solutions that are not immediately visible, enhancing efficiency in mathematical problem-solving.

Derivative of Logarithmic Functions

The derivative of logarithmic functions is a key concept in calculus. For a function like \( f(x) = \ln(u(x)) \), where \( u(x) \) is a differentiable function, the derivative is given by:

• \( \frac{d}{dx} \ln(u(x)) = \frac{1}{u(x)} \cdot u'(x) \).

This is known as the chain rule, used to differentiate composite functions.In our specific case, we differentiate \( f(x) = \ln(\sin(x)) \). By applying the chain rule,

• \( f'(x) = \frac{1}{\sin(x)} \cdot \cos(x) = \cot(x) \).

This derivative is then used in the arc length formula we discussed earlier. Derivatives of logarithmic functions are widely used in real-world scenarios such as growth models, economic forecasts, and anywhere the relationship is multiplicatively proportional.
Understanding how to differentiate these functions is crucial for solving many complex calculus problems efficiently.