Problem 55 - Calculus Early Transcendentals: Pearson New International Edition

Question

Find the length of the curve \(y=\ln (\cos x)\) between \(x=0\) and \(x=\pi / 4\).

Step-by-Step Solution

Answer

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

Step 1: Formula for Arc Length of a Curve

The formula for finding the length of a curve given by a function \( y = f(x) \) from \( x = a \) to \( x = b \) is \[L = \int_a^b \sqrt{1 + \left( \frac{dy}{dx} \right)^2} \, dx.\] We will apply this formula to our specific function \( y = \ln(\cos x) \).

Step 2: Find the Derivative \(\frac{dy}{dx}\)

To find \( \frac{dy}{dx} \), we differentiate \( y = \ln(\cos x) \) using the chain rule. The derivative is \[\frac{dy}{dx} = \frac{d}{dx}[\ln(\cos x)] = \frac{-\sin x}{\cos x} = -\tan x.\]

Step 3: Square the Derivative

Next, calculate \( \left( \frac{dy}{dx} \right)^2 = (-\tan x)^2 = \tan^2 x \).

Step 4: Set Up the Integral for Arc Length

Substitute \( \left( \frac{dy}{dx} \right)^2 \) into the arc length formula:\[L = \int_0^{\pi/4} \sqrt{1 + \tan^2 x} \, dx.\] Simplify inside the integral using the trigonometric identity \(1 + \tan^2 x = \sec^2 x\).Then,\[L = \int_0^{\pi/4} \sqrt{\sec^2 x} \, dx = \int_0^{\pi/4} \sec x \, dx.\]

Step 5: Evaluate the Integral

The integral of \( \sec x \) is \( \ln |\sec x + \tan x| + C \). So, we evaluate:\[L = \left[ \ln |\sec x + \tan x| \right]_0^{\pi/4}.\] At \( x = \pi/4 \), \( \sec (\pi/4) = \sqrt{2} \) and \( \tan (\pi/4) = 1 \).Thus, \( \ln |\sec (\pi/4) + \tan (\pi/4)| = \ln (\sqrt{2} + 1) \).At \( x = 0 \), \( \sec(0) = 1 \) and \( \tan(0) = 0 \).Thus, \( \ln |1 + 0| = 0 \).Therefore, \[L = \ln(\sqrt{2} + 1) - 0 = \ln(\sqrt{2} + 1).\]

Step 6: Conclusion

The length of the curve \( y = \ln(\cos x) \) from \( x = 0 \) to \( x = \pi/4 \) is \( \ln(\sqrt{2} + 1) \).

Key Concepts

Chain Rule

The chain rule is a fundamental concept in calculus, allowing us to differentiate composite functions. When dealing with a function like \( y = \ln(\cos x) \), we view it as a composition of the natural logarithm and the cosine function. The chain rule states that if you have a function \( y = f(g(x)) \), the derivative \( \frac{dy}{dx} \) is given by \( f'(g(x)) \cdot g'(x) \).
This means we first differentiate the outer function, leaving the inner function unchanged, and then multiply by the derivative of the inner function.

In our case, for \( y = \ln(\cos x) \), the outer function is \( \ln(u) \) with \( u = \cos x \). The derivative of \( \ln(u) \) with respect to \( u \) is \( \frac{1}{u} \). The inner function \( u = \cos x \) has a derivative of \(-\sin x \).

• Apply the chain rule: \( \frac{dy}{dx} = \frac{1}{\cos x} \cdot (-\sin x) = -\tan x \).

This simplification showcases the power and utility of the chain rule in differentiating complex functions.

Trigonometric Identity

Trigonometric identities are equations involving trigonometric functions that are true for all values of the variables where the functions are defined. In our exercise, one crucial trigonometric identity is \( 1 + \tan^2 x = \sec^2 x \). This identity comes from the Pythagorean identity and is key in simplifying complex calculations involving tangent and secant functions.

• The identity helps to simplify \( \sqrt{1 + \tan^2 x} \) into \( \sqrt{\sec^2 x} \), which equals \( \sec x \).
• Thus, the integral \( \int \sqrt{1 + \tan^2 x} \, dx \) changes to \( \int \sec x \, dx \).

This simplification makes evaluating the integral more straightforward, showcasing how knowing and applying trigonometric identities can greatly simplify calculations.

Integral Calculus

Integral calculus is about finding areas under curves and solving related problems. The arc length of a curve \( y = f(x) \) over an interval \([a, b]\) uses the formula:

• \[L = \int_a^b \sqrt{1 + \left( \frac{dy}{dx} \right)^2} \, dx\]

In this problem, the derivative \( \frac{dy}{dx} \) is \(-\tan x \). Following the steps, we compute \( \left( \frac{dy}{dx} \right)^2 = \tan^2 x \) leading to a simplified integral for arc length:

• \[L = \int_0^{\pi/4} \sec x \, dx\]

To integrate \( \sec x \), we use its antiderivative \( \ln |\sec x + \tan x| + C \). Applying the limits of integration, we evaluate the definite integral to find the arc length. The calculated arc length is \( \ln(\sqrt{2} + 1) \).
This shows how integral calculus allows us to compute measurements like arc length by finding and evaluating the appropriate antiderivatives.

Logarithmic Function

Logarithmic functions involve the natural logarithm, denoted as \( \ln(x) \), which is the inverse of the exponential function. These functions are useful in various calculus and applied mathematics problems due to their unique properties.

For the problem at hand, \( y = \ln(\cos x) \) involves applying these logarithmic properties in differentiation and integration processes. When we differentiate \( \ln(f(x)) \), we use the formula \( \frac{1}{f(x)} \cdot f'(x) \), which is directly applied in the chain rule strategy.
This function's graph mirrors another crucial aspect of logarithmic functions—understanding how they behave and transform depending on their arguments. In integral evaluations, like converting \( \ln(\sqrt{2} + 1) \), understanding logarithmic operations becomes essential.
By incorporating the logarithmic property \( \ln(a \cdot b) = \ln(a) + \ln(b) \) and similar identities, we manage transformations and simplifications that occur in these evaluations, clarifying how logarithms play at the center of calculus applications.