Problem 38 - Calculus Single Variable

Question

Calculate the arc length \(L\) of the graph of the given equation. $$ y=\ln \left(x^{2}-1\right) \quad \sqrt{2} \leq x \leq \sqrt{3}. $$

Step-by-Step Solution

Answer

The arc length is \( \sqrt{3} - \sqrt{2} + 2 \ln 2 \).

Step 1: Write the Arc Length Formula

The arc length of a curve described by a function \( y = f(x) \) from \( x = a \) to \( x = b \) is given by the formula \[ L = \int_{a}^{b} \sqrt{1 + \left( \frac{dy}{dx} \right)^2 } \, dx. \] In this problem, \( a = \sqrt{2} \) and \( b = \sqrt{3} \).

Step 2: Differentiate the Function

We know \( y = \ln(x^2 - 1) \). To find the derivative \( \frac{dy}{dx} \), apply the chain rule:\[ \frac{dy}{dx} = \frac{1}{x^2 - 1} \cdot 2x = \frac{2x}{x^2 - 1}. \]

Step 3: Compute the Expression Inside the Integral

Substitute \( \frac{dy}{dx} = \frac{2x}{x^2 - 1} \) into the arc length formula:\[ 1 + \left( \frac{dy}{dx} \right)^2 = 1 + \left( \frac{2x}{x^2 - 1} \right)^2 = 1 + \frac{4x^2}{(x^2 - 1)^2}. \]

Step 4: Simplify the Expression

Simplify \( 1 + \frac{4x^2}{(x^2 - 1)^2} \):\[ 1 + \frac{4x^2}{(x^2 - 1)^2} = \frac{(x^2 - 1)^2 + 4x^2}{(x^2 - 1)^2}. \]Expanding and simplifying gives:\[ (x^2 - 1)^2 + 4x^2 = x^4 - 2x^2 + 1 + 4x^2 = x^4 + 2x^2 + 1 = (x^2 + 1)^2. \]Thus,\[ 1 + \left( \frac{dy}{dx} \right)^2 = \frac{(x^2 + 1)^2}{(x^2 - 1)^2}. \]

Step 5: Evaluate the Integral

Substitute back into the arc length integral:\[ L = \int_{\sqrt{2}}^{\sqrt{3}} \sqrt{ \frac{(x^2 + 1)^2}{(x^2 - 1)^2} } \, dx = \int_{\sqrt{2}}^{\sqrt{3}} \frac{x^2 + 1}{x^2 - 1} \, dx. \]

Step 6: Integrate the Function

To compute \( \int \frac{x^2 + 1}{x^2 - 1} \, dx \), decompose it:\[ \frac{x^2 + 1}{x^2 - 1} = 1 + \frac{2}{x^2 - 1}. \]Thus,\[ L = \int_{\sqrt{2}}^{\sqrt{3}} \left( 1 + \frac{2}{x^2 - 1} \right) \, dx. \]Integrating term by term:- \( \int 1 \, dx = x \).- \( \int \frac{2}{x^2 - 1} \, dx = \ln |x^2 - 1| \).So, \( L = \left[ x + 2 \ln |x^2 - 1| \right]_{\sqrt{2}}^{\sqrt{3}}. \)

Step 7: Calculate the Arc Length

Substitute the limits into the integrated function:\[ L = \left( \sqrt{3} + 2 \ln \left| 2 \right| \right) - \left( \sqrt{2} + 2 \ln \left| 1 \right| \right). \]Since \( \ln |1| = 0 \), this reduces to:\[ L = \sqrt{3} - \sqrt{2} + 2 \ln 2. \]

Key Concepts

Integration

Integration is like adding up tiny slices of an area to find a total. This process is crucial when calculating the arc length because you need to sum up the infinitesimal segments along a curve. In the context of arc length, integration helps determine the distance along a smooth, curved line between two points.

For our problem of finding the arc length of the function \( y = \ln(x^2 - 1) \) from \( x = \sqrt{2} \) to \( x = \sqrt{3} \), we use the arc length formula:

• \[ L = \int_{a}^{b} \sqrt{1 + \left( \frac{dy}{dx} \right)^2 } \, dx \]
• Here, integration combines all the calculated \'slices\' from \( x = \sqrt{2} \) to \( x = \sqrt{3} \), delivering the total arc length.

The integral setup encompasses both the function's geometry and its change rate along the x-axis, captured by \( \frac{dy}{dx} \). This way, integration not only sums up values but also respects the curve's slope, ensuring accuracy in representing the actual arc length.

Remember, it's theintegration that binds math to the real world by connecting individual moments to find totals, just like joining short segments to get the length of a curve.

Differentiation

Differentiation is the process of finding a derivative, which tells us how much a function changes at any point—the function's slope. In solving for arc length, differentiation is integral because it helps determine how steep or flat the curve is at each segment.

For our example, given the function \( y = \ln(x^2 - 1) \), we must differentiate to obtain \( \frac{dy}{dx} \):

• Using the chain rule, \( y = \ln(u) \) where \( u = x^2 - 1 \).
• The derivative is \( \frac{dy}{dx} = \frac{1}{u} \cdot \frac{du}{dx} \).
• Calculating \( \frac{du}{dx} = 2x \), thus \( \frac{dy}{dx} = \frac{2x}{x^2 - 1} \).

This derivative \( \frac{dy}{dx} \), expresses the rate of change of \( y \) with respect to \( x \), crucial for understanding the shape and trajectory of the curve at any point between \( x = \sqrt{2} \) and \( x = \sqrt{3} \).

Differentiation acts like a zoom lens, revealing not just where you are on a curve, but how the path is bending at every step.

Chain Rule

The chain rule is a key principle in calculus for handling complex functions by breaking them into simpler ones. It guides us in differentiating a composition of functions. When you have a function like \( y = \ln(x^2 - 1) \), it's composed of the natural logarithm and a quadratic function.

Applying the chain rule involves these steps:

• Identify the inner function \( u = x^2 - 1 \) and the outer function \( y = \ln(u) \).
• Differentiate the outer function with respect to \( u \), \( \frac{dy}{du} = \frac{1}{u} \).
• Find the derivative of the inner function, \( \frac{du}{dx} = 2x \).
• Combine the derivatives using the chain rule: \( \frac{dy}{dx} = \frac{dy}{du} \times \frac{du}{dx} \).
• This results in \( \frac{dy}{dx} = \frac{2x}{x^2 - 1} \).

The chain rule essentially links the changes in the nested parts of a function to derive the overall change concerning \( x \). It's like reading a map of dependencies among function layers, keeping track of how adjusting one aspect influences the entire function.

By mastering the chain rule, we can confidently tackle complex derivatives, unlocking deeper insights into function behavior.