Question

[T] Find the arc length of \(\ln x\) from \(x=1\) to \(x=2\).

Short answer

The arc length is \( \ln\left(\frac{\sqrt{5} + 2}{\sqrt{2} + 1}\right) \).

Step-by-step solution

Understand the Formula for Arc Length

The formula to find the arc length of a function \( y = f(x) \) from \( x = a \) to \( x = b \) is \( L = \int_a^b \sqrt{1 + \left( \frac{dy}{dx} \right)^2} \, dx \). For this problem, the function is \( y = \ln x \).

Differentiate the Function

Find the derivative of \( y = \ln x \). The derivative is \( \frac{dy}{dx} = \frac{1}{x} \).

Substitute and Simplify

Substitute \( \frac{dy}{dx} = \frac{1}{x} \) into the arc length formula: \( L = \int_1^2 \sqrt{1 + \left( \frac{1}{x} \right)^2} \, dx \). Simplify inside the square root: \( 1 + \left( \frac{1}{x} \right)^2 = 1 + \frac{1}{x^2} = \frac{x^2 + 1}{x^2} \).

Integrate

Substitute the simplification into the integral: \( L = \int_1^2 \sqrt{\frac{x^2+1}{x^2}} \, dx = \int_1^2 \frac{\sqrt{x^2+1}}{x} \, dx \). Use a suitable substitution or integration technique to evaluate the integral. A common technique for forms like this is trigonometric substitution.

Use Trigonometric Substitution

Let \( x = \tan(\theta) \), then \( dx = \sec^2(\theta) \, d\theta \) and \( \sqrt{x^2 + 1} = \sec(\theta) \). Substitute and change the limits: When \( x = 1, \theta = \frac{\pi}{4} \); when \( x = 2, \theta = \arctan(2) \). The integral becomes \( \int_{\frac{\pi}{4}}^{\arctan(2)} \sec(\theta) \, d\theta \).

Integrate Sec(θ)

The integral of \( \sec(\theta) \) is \( \ln | \sec(\theta) + \tan(\theta) | + C \). Evaluate this from \( \frac{\pi}{4} \) to \( \arctan(2) \).

Evaluate the Integral

Substitute the limits into the result from Step 6: - When \( \theta = \arctan(2) \), \( \sec(\theta) + \tan(\theta) = \sqrt{1 + 4} + 2 = \sqrt{5} + 2 \).- When \( \theta = \frac{\pi}{4} \), \( \sec(\theta) + \tan(\theta) = \sqrt{2} + 1 \).The result is \( \ln(\sqrt{5} + 2) - \ln(\sqrt{2} + 1) \).

Simplify the Final Answer

Combine the logarithms using properties of logarithms: \( \ln(\sqrt{5} + 2) - \ln(\sqrt{2} + 1) = \ln\left(\frac{\sqrt{5} + 2}{\sqrt{2} + 1}\right) \). This is the final answer for the arc length.

Key concepts

Integration Techniques

Integration techniques are methods used to evaluate integrals, which are essential in finding values like area and arc length. In this problem, you need to integrate to determine the arc length of the logarithmic function from one point to another. The arc length formula involves the integral of the square root of a function and its derivative. Different techniques can simplify such integrals, including:

• Substitution method: Simplifies integrals by changing the variable of integration.
• Integration by parts: Breaks down products of functions into easier parts.
• Partial fraction decomposition: Decomposes complex fractions into simpler parts.

Each technique has its use case and choosing the right one can significantly ease the solving process. In this exercise, trigonometric substitution is a key integration technique used to handle the square root expression.

Trigonometric Substitution

Trigonometric substitution is a powerful technique for evaluating integrals involving square roots. It is especially useful when the integrals contain forms like \( \sqrt{x^2 + a^2} \). By substituting a trigonometric function, these forms can become easier to integrate.
In this exercise, we use the substitution \( x = \tan(\theta) \), which converts \( \sqrt{x^2 + 1} \) to \( \sec(\theta) \). This transformation simplifies subsequent steps, as trigonometric functions often have readily available integrals.

• First, identify the substitution: \( x = \tan(\theta) \).
• Convert \( dx \) using \( dx = \sec^2(\theta) \, d\theta \).
• Change the limits of integration according to this substitution.

Once these changes are made, the integral becomes a simpler function of \( \theta \), making it straightforward to integrate.

Differentiation

Differentiation is the process of finding the derivative of a function, which represents the rate of change. In the context of finding arc length, the derivative is used to measure the slope of the curve at any given point.
For the function \( y = \ln x \), the derivative \( \frac{dy}{dx} \) is \( \frac{1}{x} \). This derivative is essential as it is incorporated into the arc length formula.

• Start by differentiating \( y = \ln x \) to get \( \frac{dy}{dx} = \frac{1}{x} \).
• Use this derivative in the arc length formula, \( L = \int_a^b \sqrt{1 + \left( \frac{dy}{dx} \right)^2} \, dx \).
• Calculate \( 1 + \left( \frac{1}{x} \right)^2 \) to prepare the integrand.

Understanding how to differentiate correctly is crucial as it impacts the whole integration process that follows.

Logarithmic Functions

Logarithmic functions, such as \( y = \ln x \), are inverse operations to exponentiation and show how many times a base number is multiplied to get another number. These functions are significant in various mathematical contexts, including calculus and arc length calculations.
In this problem, the length of the arc is determined for \( y = \ln x \) over a specific interval.

• Logarithmic functions have unique properties, like the derivative \( \frac{dy}{dx} = \frac{1}{x} \) for \( y = \ln x \).
• Their graphs typically have a shape that rises without boundary as \( x \) increases.
• Calculating the arc length of such functions involves both their properties and derivatives.

Logarithmic functions showcase intriguing mathematical behaviors and their derivatives simplify finding integrals associated with their curves.