91Ó°ÊÓ

The arc length formula is a method in calculus used to determine the distance along a curve between two points. To find the arc length of a curve described by the function \[ y = f(x) \] from \( x = a \) to \( x = b \), we use the formula:

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

This formula incorporates a derivative, \( \frac{dy}{dx} \), which represents the slope of the curve. The formula basically measures small distances along the curve by computing the hypotenuse of small right triangles formed by horizontal and vertical changes in the function. Then, it integrates these small lengths to obtain the total arc length. It is crucial to understand that this formula can be applied to different functions as long as we can find their derivatives.

Finding the derivative is a fundamental skill in calculus. Derivatives tell us about the rate of change of a function, or its slope, at any given point. For the function \( y = \ln(\cos x) \), we need to differentiate using the chain rule. The chain rule states that if you have a composite function, the derivative is the derivative of the outer function times the derivative of the inner function.For our function:

• \( y = \ln(\cos x) \)
• The derivative is \( \frac{dy}{dx} = \frac{1}{\cos x}(-\sin x) = -\tan x \)

Here, we used trigonometric identities, where the derivative of \( \ln u \) gives \( \frac{1}{u} \cdot \frac{du}{dx} \), and the chain automatically gives us \( -\sin x \) from \( \cos x \). Getting comfortable with differentiating various types of functions, particularly ones involving logarithms and trigonometric identities, is key for solving many calculus problems.

Integrals are used to accumulate quantities over an interval, which in the case of arc length, helps calculate the total length along a curve. In our example, after substituting the derivative into the arc length formula, we simplify it to

• \[ L = \int_0^{\pi/3} \sqrt{1 + (-\tan x)^2} \, dx \]
• Which becomes \[ L = \int_0^{\pi/3} \sqrt{\sec^2 x} \, dx \]
• Resulting in \[ L = \int_0^{\pi/3} |\sec x| \, dx \]

The integral then requires evaluation over the interval from \( 0 \) to \( \pi/3 \). For \( \sec x \), the antiderivative is known and is expressed as \( \ln |\sec x + \tan x| \). The boundaries of the interval, \( 0 \) to \( \pi/3 \), are used to evaluate the definite integral, revealing the length of the curve over the interval.

Trigonometric identities are essential tools in calculus for simplifying expressions and solving integrals. In this solution, one identity stands out:

• \( 1 + \tan^2 x = \sec^2 x \)

This identity transforms complicated expressions into simpler ones, enabling us to solve integrals more efficiently. By substituting \( 1 + \tan^2 x \) with \( \sec^2 x \), the integral becomes manageable. Always remember that trigonometric identities can change the form of an expression, allowing for easier computation. Integrals involving expressions like \( \sec x \) often appear, and knowing its relationship with \( \tan x \) can greatly assist in evaluating these integrals correctly.