Question

Consider the curve \(y=\ln x\) a. Find the length of the curve from \(x=1\) to \(x=a\) and call it \(L(a) .\) (Hint: The change of variables \(u=\sqrt{x^{2}+1}\) allows evaluation by partial fractions.) b. Graph \(L(a)\) c. As \(a\) increases, \(L(a)\) increases as what power of \(a ?\)

Short answer

Question: Determine the expression for the length of the curve \(y = \ln x\) from \(x = 1\) to \(x = a\), and describe how the curve's length increases as \(a\) increases. Answer: The length of the curve, denoted as \(L(a)\), is given by the expression: $$L(a) = \frac{1}{2}\left(a^2 + 1\right) - \frac{1}{2}$$ As \(a\) increases, the length of the curve \(L(a)\) increases as the square of \(a\).

Step-by-step solution

Arc Length Formula

To find the length of the curve from \(x=1\) to \(x=a\), we will use the arc length formula given by: $$L(a) = \int_{1}^{a} \sqrt{1 + \left(\frac{dy}{dx}\right)^2}\ dx$$ First, we need to find the derivative of \(y = \ln x\) with respect to \(x\).

Finding the Derivative of \(y\) with respect to \(x\)

Take the derivative of \(y = \ln x\) with respect to \(x\): $$\frac{dy}{dx} = \frac{1}{x}$$

Substitute the Derivative into the Arc Length Formula

Plug \(\frac{dy}{dx} = \frac{1}{x}\) into the arc length formula: $$L(a) = \int_{1}^{a} \sqrt{1 + \left(\frac{1}{x}\right)^2}\ dx$$

Change of Variables

Use the suggested change of variables, letting \(u = \sqrt{x^2 + 1}\), and find the substitution we need to make to replace \(du\) in terms of \(dx\): $$u^2 = x^2 + 1 \Rightarrow 2u \frac{du}{dx} = 2x \Rightarrow \frac{du}{dx} = \frac{x}{u}$$ Now, let's find \(x\) in terms of \(u\) to substitute into our integral equation: $$x^2 = u^2 - 1 \Rightarrow x = \sqrt{u^2 - 1}$$ Now, we can substitute \(x\) and \(\frac{du}{dx}\) into our integral: $$L(a) = \int_{1}^{\sqrt{a^2 + 1}} u\ du$$

Evaluate the Integral

Evaluate the integral to find \(L(a)\): $$L(a) = \frac{1}{2}u^2 \Big|_1^{\sqrt{a^2 + 1}} = \frac{1}{2}\left(\sqrt{a^2 + 1}\right)^2 - \frac{1}{2} = \frac{1}{2}\left(a^2 + 1\right) - \frac{1}{2}$$ Now, we have the expression for \(L(a)\): $$L(a) = \frac{1}{2}\left(a^2 + 1\right) - \frac{1}{2}$$

Graph \(L(a)\)

Graph the function \(L(a) = \frac{1}{2}\left(a^2 + 1\right) - \frac{1}{2}\) to visualize how \(L(a)\) changes with respect to \(a\). The graph is a quadratic function that is increasing with respect to \(a\).

Determine the Power of \(a\)

Looking at the expression for \(L(a)\), the highest power of \(a\) is 2. Therefore, as \(a\) increases, \(L(a)\) increases as the square of \(a\). The final answer is: a. \(L(a) = \frac{1}{2}\left(a^2 + 1\right) - \frac{1}{2}\) b. (Graph of \(L(a)\) as a quadratic function) c. \(L(a)\) increases as the square of \(a\).

Key concepts

Integral Calculus

Integral Calculus is an essential part of Calculus that deals with the concept of integration, which is the reverse process of differentiation. It helps in finding areas under curves, volumes, central points, and many more applications. In this problem, we are interested in finding the arc length of the curve described by the function \(y = \ln x\). The arc length \(L(a)\) from \(x=1\) to \(x=a\) can be calculated using the integration form of the arc length formula:

• Start with the formula: \[L(a) = \int_{1}^{a} \sqrt{1 + \left(\frac{dy}{dx}\right)^2}\ dx\]
• The derivative \(\frac{dy}{dx}\) needs to be computed first to substitute into the formula.
• The variable \(x\) is integrated over the interval, adding small increments of arc length to find the total length from 1 to \(a\).

Integration is the key tool here. It transforms the function and derivatives into a calculable form, giving us the length of the curve after evaluation.

Derivative of a Function

The Derivative of a function measures how the function's output changes in response to changes in its input. In simple terms, it provides the slope or steepness of the function at a given point. For this problem, we deal with the natural logarithm function \(y = \ln x\). Here's how we deal with it:

• To find \(\frac{dy}{dx}\), take the derivative of \(y = \ln x\), which results in \(\frac{1}{x}\).
• This derivative expresses how rapidly \(y\) changes as \(x\) varies.

The derivative \(\frac{1}{x}\) is critical because it specifies part of the expression under the integral sign, affecting how the arc's length is calculated. By knowing this rate of change, we essentially track how lengths differ as we traverse along the curve as \(x\) increases.

Change of Variables

The Change of Variables technique is a powerful method in integration that simplifies complex integrals into more manageable forms. In this exercise, we were guided to use a change of variable, \(u = \sqrt{x^2 + 1}\), to facilitate the calculation. Here's a breakdown:

• By setting \(u = \sqrt{x^2 + 1}\), we essentially transform the problem into a new variable.
• This allows us to express expressions in terms of \(u\) rather than \(x\), often simplifying the integral.
• Convert differential components: From \(du = \frac{x}{u} dx\), rearrange to substitute \(dx\).

This transformation acts like changing perspectives, making it easier to evaluate the integral. The substitution streamlines the process of integrating, turning an otherwise complicated integral into a simpler, more straightforward one.

Partial Fractions

Partial Fractions is a method widely used to simplify complex rational expressions, especially those encountered in integration. In this particular problem, after changing the variables, the integral becomes quite straightforward and doesn't need partial fractions per se. Yet, understanding the concept can be insightful:

• Partial fractions involve breaking up a complex fraction into simpler fractions that are much easier to integrate.
• This technique is essential, especially when dealing with integrals resulting from polynomial expressions in the denominator.

For our curve \(y = \ln x\), the integral became manageable without needing partial fractions, primarily due to the effective change of variables. Nevertheless, being conversant with partial fractions is crucial as they are usually a staple in integral calculus for decomposing fractions in integration problems.