Problem 36 - Calculus Single Variable

Question

Calculate the arc length \(L\) of the graph of the given equation. $$ y=\ln (x) \quad 1 \leq x \leq e. $$

Step-by-Step Solution

Answer

The arc length \( L = \sqrt{e^2 + 1} - \sqrt{2} \).

Step 1: Understand the Arc Length Formula

The arc length of a graph from \(x = a\) to \(x = b\) is given by the formula \( L = \int_a^b \sqrt{1 + \left(\frac{dy}{dx}\right)^2} \, dx \). This formula requires us to find the derivative \( \frac{dy}{dx} \) and then integrate over the interval [a, b].

Step 2: Compute the Derivative

For the function \( y = \ln(x) \), find the derivative: \( \frac{dy}{dx} = \frac{1}{x} \). This derivative will be used in the arc length formula.

Step 3: Substitute into Arc Length Formula

Substitute \( \frac{dy}{dx} = \frac{1}{x} \) into the formula, yielding \( L = \int_1^e \sqrt{1 + \left(\frac{1}{x}\right)^2} \, dx \). Simplifying inside the square root gives \( L = \int_1^e \sqrt{1 + \frac{1}{x^2}} \, dx \).

Step 4: Simplify the Expression under the Square Root

Calculate \( 1 + \frac{1}{x^2} = \frac{x^2 + 1}{x^2} \). The square root becomes \( \sqrt{\frac{x^2 + 1}{x^2}} = \frac{\sqrt{x^2 + 1}}{x} \). This simplifies our integral to \( L = \int_1^e \frac{\sqrt{x^2 + 1}}{x} \, dx \).

Step 5: Make a Substitution to Solve the Integral

Use the substitution \( u = x^2 + 1 \), which implies \( du = 2x \, dx \), or \( x \, dx = \frac{1}{2} du \). Adjust limits: when \( x = 1, \, u = 2 \); when \( x = e, \, u = e^2 + 1 \). This substitution transforms \( L = \int_2^{e^2 + 1} \frac{1}{2} u^{-\frac{1}{2}} \, du \).

Step 6: Integrate with Respect to New Variable

Evaluate the integral \( L = \frac{1}{2} \int_2^{e^2 + 1} u^{-\frac{1}{2}} \, du \). Integrate to find \( \frac{1}{2} \left[ 2 u^{\frac{1}{2}} \right]_2^{e^2 + 1} = \left[ u^{\frac{1}{2}} \right]_2^{e^2 + 1} \).

Step 7: Evaluate the Definite Integral

Substitute back the limits: \( L = \left[(e^2 + 1)^{1/2} - 2^{1/2}\right] \). Simplify this expression to find the arc length.

Key Concepts

Arc Length Formula

The arc length formula is an essential tool in calculus that allows us to determine the length of a curve defined by a function over a certain interval. For a function \( y = f(x) \) that is smooth over the interval \([a, b]\), the arc length \( L \) is given by the equation: \[ L = \int_a^b \sqrt{1 + \left(\frac{dy}{dx}\right)^2} \, dx \] This formula calculates the distance along the curve from point \( x = a \) to point \( x = b \). Importantly, you will need the derivative \( \frac{dy}{dx} \) of the function to use this formula effectively. The arc length formula essentially sums up all the tiny line segments that approximate the curve, factoring in the slope of each segment using the derivative.

Integration Techniques

To solve problems involving arc length, you usually need to apply integration techniques. Integration is the process of finding the integral, which is the reverse operation of differentiation. When tackling the arc length formula:

• First, identify the function and its derivative.
• Substitute the derivative into the formula, creating the integral expression.
• Look for ways to simplify the integral. You might factor expressions, use substitutions, or apply partial fraction decomposition.

In the example of \( y = \ln(x) \), once substituting the derivative into the arc length formula, you end up with:\[ L = \int_1^e \frac{\sqrt{x^2 + 1}}{x} \, dx \] This requires some manipulation before proceeding, including simplification and substitution, to evaluate the integral effectively.

Chain Rule Substitution

A powerful tool for evaluating integrals is substitution, particularly the chain rule substitution technique. This technique involves changing variables to simplify the integration process. In the arc length problem for \( y = \ln(x) \), we used the substitution: \( u = x^2 + 1 \) This substitution helps by transforming a complicated argument inside the square root into a simpler form. Here’s how you can apply this:

• Find the derivative \( du = 2x \, dx \) and manipulate it so that \( x \, dx = \frac{1}{2} du \).
• Change the limits of integration to reflect the new variable. For example, when \( x = 1 \) changes to \( u = 2 \), and when \( x = e \) changes to \( u = e^2 + 1 \).

After substitution, the integral looks like:\[ L = \int_2^{e^2 + 1} \frac{1}{2} u^{-\frac{1}{2}} \, du \] This makes the integral more straightforward to solve.

Calculus Problem Solving

Calculus problem solving is about applying your foundational calculus skills to tackle complex problems effectively. Every calculus problem, like finding the arc length, requires methodical steps:

• Understand the problem: Grasp what is being asked. Identify known variables and what you need to find.
• Set up equations: Use formulas, like the arc length formula, with the correct expressions and substitutions.
• Apply calculus operations: Derive the necessary derivatives and integrate using suitable techniques, like the chain rule substitution.
• Evaluate and Simplify: Substitute limits and simplify expressions. In our arc length problem, we simplified to \( [ (e^2 + 1)^{1/2} - 2^{1/2} ] \).
• Verify your solution: Check the process and recalibrate if necessary to ensure that the solution is correct.

Thus, mastering these steps allows you to tackle calculus problems confidently, transforming challenging questions into manageable tasks.