Question

In Exercises \(3-14,\) find the arc length of the graph of the function over the indicated interval. $$ y=\ln (\cos x), \quad\left[0, \frac{\pi}{3}\right] $$

Short answer

The arc length of the function \(y=\ln (\cos x)\) over the interval \([0, \frac{\pi}{3}]\) is \(L= \ln(3\sqrt{3})\).

Step-by-step solution

Find the Derivative

The derivative of the function \(y=\ln (\cos(x))\) is given by the chain rule: \(y' = \frac{-\sin(x)}{\cos(x)} = -\tan(x)\).

Apply the Arc Length Formula

Input the derivative into the formula for the arc length: \(L = \int_a^b{\sqrt{1+(f'(x))^2}}dx = \int_0^{\pi/3}{\sqrt{1+(-\tan(x))^2}}dx = \int_0^{\pi/3}{\sqrt{1+\tan^2(x)}}dx\).

Simplify the Integral

Utilize the Pythagorean identity \(\tan^2(x) + 1 = \sec^2(x)\) to simplify the square root: \(L = \int_0^{\pi/3}{\sqrt{\sec^2(x)}}dx = \int_0^{\pi/3}{\sec(x)}dx\).

Solve the Integral

The integral of \(\sec(x)\) can be found in a table of standard integrals: \(\int{\sec(x)}dx = \ln |\sec(x) + \tan(x)|\). Evaluate this from 0 to \(\pi/3\) to get: \(\ln |\sec(\frac{\pi}{3}) + \tan(\frac{\pi}{3})| - \ln |\sec(0) + \tan(0)| = \ln(2\sqrt{3} +\sqrt{3}) - \ln(1)\).

Find the Final Answer

Take the difference of the logarithms using the property \(\ln a - \ln b = \ln(a/b)\) to find: \(L= \ln( \frac{2\sqrt{3} +\sqrt{3}}{1}) = \ln(3\sqrt{3})\).

Key concepts

Chain Rule

When you are asked to find the arc length of a function, particularly if it involves a composite function like \(y=(\cos x)\), understanding the chain rule is crucial. The chain rule is a fundamental principle in calculus used to differentiate composite functions. In other words, if you have a function that is made up of other functions, the chain rule helps you find the derivative of this combination.

When applied to our function \(y=(\cos x)\), we focus on differentiating the inner function \(\cos x\) and the outer function \((\cdot)\). The derivative of the outer function (with respect to its inside) is \(\frac{1}{\cdot}\), while the derivative of \(\cos x\) with respect to \(x\) is \(- x\). The chain rule tells us to multiply these derivatives for our final derivative, giving us \(y' = \frac{-(x)}{\cos x} = -\tan(x)\).

This step is critical because getting the derivative correct is key to accurately determining arc length using the subsequent steps in the integral calculus process.

Integral Calculus

Integral calculus is a branch of mathematics focused on finding the total size, length, area, or volume of objects. When we look at finding the arc length of the function \(y=(\cos x)\), integral calculus is used to sum up an infinite number of infinitesimally small pieces along the curve of the function. The arc length formula for a function \(f(x)\) from point \(a\) to point \(b\) is expressed as \(L = _a^b{\sqrt{1+(f'(x))^2}}dx\).

For our function, we first plug in the derivative to the arc length formula: \(L = _0^{\pi/3}{\sqrt{1+(-\tan(x))^2}}dx = _0^{\pi/3}{\sqrt{1+\tan^2(x)}}dx\). The integral calculus then involves simplifying the expression under the square root using trigonometric identities and finding the integral of the simplified function, which leads to finding the exact length of the arc.

Pythagorean Identity

To simplify the arc length integral, we employ a fundamental trigonometric fact known as the Pythagorean identity. This identity relates the sides of a right triangle with its hypotenuse and states that for any angle \(x\), \(^2(x) + 1 = \sec^2(x)\).

The arc length formula involves the expression \(\sqrt{1+(\tan(x))^2}\). Applying the Pythagorean identity, we replace \(1+\tan^2(x)\) with \(\sec^2(x)\) because they are equivalent. This simplification plays a crucial role because it turns a complex square root into a manageable expression for which we have known antiderivatives. The ability to recognize and utilize such identities can transform a challenging integral into a much more straightforward problem, allowing us to proceed to the final calculation of the arc length.