Problem 63 - Calculus Early Transcendentals

Question

Find the length of the curve \(y=\ln (\sec x),\) for \(0 \leq x \leq \pi / 4\).

Step-by-Step Solution

Answer

Question: Determine the length of the curve \(y = \ln(\sec x)\) on the interval \(0 \leq x \leq \pi / 4\). Answer: The length of the curve \(y = \ln(\sec x)\) for \(0 \leq x \leq \pi / 4\) is \(L = \ln(\sqrt{2} + 1)\).

Step 1: Find the derivative of the function

Given the function \(y = \ln (\sec x),\) first, we need to find its derivative with respect to \(x\).Using the chain rule for the logarithm and the differentiation of sec: $$\frac{dy}{dx} = \frac{1}{\sec x} \cdot \frac{d(\sec x)}{dx} = \frac{1}{\sec x} \cdot (\sec x \tan x)$$ Now simplify the expression: $$\frac{dy}{dx} = \tan x$$

Step 2: Square the derivative and add 1

Now, we need to find \(1+[\tan x]^2\). Notice that: $$1+\tan^2 x = \sec^2 x$$

Step 3: Compute the integral

The arc length formula is given by: $$L = \int_{a}^{b} \sqrt{1 + [f'(x)]^2} dx$$ Thus, we need to compute the integral: $$L = \int_{0}^{\pi/4} \sqrt{\sec^2 x} dx$$ The square root of \(\sec^2 x\) is \(\sec x\). So, the integral simplifies to: $$L = \int_{0}^{\pi/4} \sec x dx$$

Step 4: Evaluate the Integral in the Given Interval

To evaluate the integral, we can use the antiderivative of \(\sec x\), which is \(\ln(\sec x + \tan x)\). Now, let's evaluate the integral in the given interval: $$L = \left[\ln(\sec x + \tan x)\right]_{0}^{\pi/4} = \ln(\sec(\pi/4) + \tan(\pi/4)) - \ln(\sec(0) + \tan(0))$$ Now, substitute the values of sec and tan: $$L = \ln(\sqrt{2} + 1) - \ln(1 + 0)$$ Thus, the length of the curve is: $$L = \ln(\sqrt{2} + 1) - \ln 1$$ Since \(\ln(1) = 0\), the final answer is: $$L = \ln(\sqrt{2} + 1)$$ So, the length of the curve \(y = \ln(\sec x),\) for \(0 \leq x \leq \pi / 4\) is \(L = \ln(\sqrt{2} + 1).\)

Key Concepts

Chain Rule

The chain rule is a fundamental principle in calculus for taking the derivative of a composition of two or more functions. When we encounter a function within another function, the chain rule tells us that the derivative is found by multiplying the derivative of the outer function by the derivative of the inner function. In other words, if we have a composition of functions such as \( f(g(x)) \), the derivative \( f'(g(x)) \times g'(x) \) encapsulates this rule.

For example, consider the function \( y = \textrm{ln}(\sec x) \) in our exercise. Here, \( \sec x \) is the inner function and the natural logarithm \( \textrm{ln} \) is the outer function. We apply the chain rule to find the derivative of \( y \) with respect to \( x \) as follows: First, we find the derivative of the outer function, which is \( 1/u \) for \( u = \sec x \) and then multiply it by the derivative of \( \sec x \) itself, which gives us \( 1 \cdot (\sec x \tan x) \), eventually simplifying to \( \tan x \) as derived in the solution.

Integral Calculus

Integral calculus is the branch of calculus that deals with finding the area under a curve or, more abstractly, finding the antiderivative or integral of a function. The process of integration is essential for calculating things like the arc length, area, volume, and other quantities that arise from accumulation. When integrating, we are essentially summing up infinitesimal pieces to find a total amount.

In the arc length problem presented, once we have the expression \( \sqrt{1 + (f'(x))^2} \) from the arc length formula, we use integral calculus to determine the total length. We take the integral from \( x = 0 \) to \( x = \pi/4 \) to accumulate the length along the curve. The antiderivative of \( \sec x \) is used here as a crucial step in evaluating the integral, leading to our final answer for the length \( L \) of the curve.

Trigonometric Functions

Trigonometric functions like sine, cosine, and tangent are periodic functions that are fundamental in the study of triangles, waves, and oscillations. These functions allow us to relate angles to side lengths in right triangles and are crucial in many areas of math and science.

In the context of our exercise, the secant function, denoted by \( \sec x \) and its properties play a vital role. The identity \( 1 + \tan^2 x = \sec^2 x \) assists in simplifying the integral. A deep understanding of trigonometric identities is crucial when working with derivatives and integrals of trigonometric functions, as they often lead to simplifications that make calculations more manageable.

Derivatives

Derivatives represent the rate at which a function is changing at any given point, essentially describing the slope or steepness of a function's graph at a specific point. They are a cornerstone of differential calculus and have myriad applications in physics, engineering, economics, and beyond.

In the given exercise, we use derivatives to determine the expression under the square root in the arc length formula. By finding \( dy/dx \), or the derivative of our function, we can then square this derivative to insert it into our arc length formula. The concept of a derivative is not only instrumental in calculating the arc length but also in countless other mathematical scenarios where change is a factor.