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Chapter 8: Problem 93

Arc Length Find the arc length of the graph of \(y=\ln (\sin x)\) from \(x=\pi / 4\) to \(x=\pi / 2\)

Short Answer

Expert verified
The arc length of the graph of \(y = \ln(\sin(x))\) from \(\pi / 4\) to \(\pi / 2\) is given by \[L = -\ln | \cot(\pi/2) + \csc(\pi/2)| + \ln | \cot(\pi/4) + \csc(\pi/4)| \].

Step by step solution

01

Compute the derivative

Differentiate the function \(y = \ln(\sin(x))\) with respect to x. The derivative of \(y\), denoted as \(y'\), is determined by the chain rule. The derivative of \(\ln\) is 1/u and the derivative of \(\sin(x)\) is \(\cos(x)\). Thus, the derivative is \(1 / \sin(x) * \cos(x)\) or \(\cot(x)\).
02

Substitute into the formula

Substitute \(y'\) into the formula for the arc length of a function on an interval \([a, b]\), given by \[L = \int_{a}^{b} \sqrt{1 + [y'(x)]^{2}} dx\] Thus, the integral to calculate is \[L = \int_{\pi/4}^{\pi/2} \sqrt{1 + (\cot(x))^2} dx\]
03

Simplify the integrand

The integral simplifies to: \[L = \int_{\pi/4}^{\pi/2} \frac{\sqrt{1 + \cot^{2}(x)}}{\sin(x)} dx\]. Rewritten as, \[L = \int_{\pi/4}^{\pi/2} \frac{1}{\sin(x)} dx.\] Now calculate the integral.
04

Evaluate the integral

Integral \(\int \frac{1}{\sin(x)} dx\) results in \(-\ln | \cot(x) + \csc(x) | \). Substituting the limits of integration, we get: \[L = -\ln | \cot(\pi/2) + \csc(\pi/2)| + \ln | \cot(\pi/4) + \csc(\pi/4)| \]].

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Understanding the Chain Rule
The chain rule is a fundamental concept in calculus used to differentiate composite functions. A composite function is a function formed by combining two or more functions, such as \(y = f(g(x))\). The chain rule helps us determine the derivative of such complex functions.

When using the chain rule, you must differentiate the outer function, keeping the inner function unchanged, and then multiply by the derivative of the inner function.

In the exercise, the function \(y = \ln(\sin(x))\) consists of two parts: the natural logarithm \((\ln)\) and the sine function \(\sin(x)\).

This is how the chain rule is applied:
  • Differentiate the outer function \(\ln(u)\) with respect to \(u\) to get \(1/u\).
  • Next, differentiate the inner function \(\sin(x)\) with respect to \(x\) to get \(\cos(x)\).
  • Multiply these derivatives together, resulting in \(1/\sin(x) \cdot \cos(x) = \cot(x)\).
By understanding the chain rule, you can simplify finding derivatives in complex problems.
Integral Calculus and Its Application
Integral calculus is essential for calculating quantities like area under a curve, total accumulation between intervals, and in this exercise, the arc length of a curve. The integral calculus focuses on antiderivatives and the definite integral, which gives the net area or accumulation over an interval.

The arc length formula involves integrating the square root of \(1 + [y'(x)]^2\) along the given interval. For our function, this means integrating from \(x=\pi/4\) to \(x=\pi/2\).

To solve for arc length:
  • Substitute the derivative \(y'(x)\) into the arc length formula: \[L = \int_{a}^{b} \sqrt{1 + [y'(x)]^{2}} \, dx\]
  • In the solution, \(y'(x) = \cot(x)\), which simplifies as you perform the integration.
  • After calculation, you complete the evaluation by plugging in the integration limits, resulting in the arc length value.
Integral calculus allows for the exact measurement of curve lengths, demonstrating its power in practical applications.
Role of Trigonometric Functions
Trigonometric functions such as sine, cosine, and tangent are key in connecting angles to ratios in right triangles, but they also play a crucial role in calculus.

In integral calculus, these functions often appear when dealing with curves that follow periodic patterns or involve circular motions.

In the given exercise, \(\sin(x)\) is utilized:
  • The term \(\sin(x)\) is part of the composite function \(\ln(\sin(x))\), requiring use of the chain rule for differentiation.
  • Its derivative, \(\cos(x)\), contributes to the simplification of the integrand during arc length calculation.
Understanding these trigonometric properties helps simplify complex integrations. Trigonometric identities often allow for straightforward manipulation and simplification of expressions within calculus concepts, making them invaluable tools in solving problems like arc length.

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Most popular questions from this chapter

$$\begin{array}{l}{\text { In Exercises } 73 \text { and } 74,(\text { a) show that the nonnegative function }} \\ {\text { is a probability density function, and (b) find } P(0 \leq x \leq 6)}\end{array}$$ $$f(t)=\left\\{\begin{array}{ll}{\frac{1}{9} e^{-t / 9},} & {t \geq 0} \\\ {0,} & {t<0}\end{array}\right.$$

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