Problem 131 - Calculus 3

Question

Find the \(x\) -coordinate at which the curvature of the curve \(y=1 / x\) is a maximum value.

Step-by-Step Solution

Answer

The maximum curvature occurs at \( x = \sqrt[4]{\frac{3}{2}} \).

Step 1: Find the First Derivative

To find the curvature, we first need the first derivative of the function. Given the function \( y = \frac{1}{x} \), we find the first derivative, \( y' \), using the power rule. The derivative is \( y' = -\frac{1}{x^2} \).

Step 2: Find the Second Derivative

Next, calculate the second derivative \( y'' \) from the first derivative \( y' = -\frac{1}{x^2} \). Differentiating again gives \( y'' = \frac{2}{x^3} \).

Step 3: Use the Curvature Formula

The formula for curvature \( K \) is \( K = \frac{|y''|}{(1 + (y')^2)^{3/2}} \). Substituting \( y' = -\frac{1}{x^2} \) and \( y'' = \frac{2}{x^3} \), the curvature \( K \) becomes \( K = \frac{2/x^3}{(1 + 1/x^4)^{3/2}} \).

Step 4: Simplify the Curvature Expression

Simplify the expression \( K = \frac{2/x^3}{(1 + 1/x^4)^{3/2}} \). Simplify the denominator:\( (1 + 1/x^4)^{3/2} = ((x^4 + 1)/x^4)^{3/2} = (x^4 + 1)^{3/2} / x^6 \, \). Thus, \( K = \frac{2 * x^3}{(x^4 + 1)^{3/2}} \).

Step 5: Find Critical Points of the Curvature Function

Set the derivative of \( K \) with respect to \( x \) to zero to find critical points. Let \( f(x) = \frac{2x^3}{(x^4 + 1)^{3/2}} \). Differentiate \( f(x) \) using the quotient rule and set it to zero to solve for \( x \).

Step 6: Solve for Maximum Curvature

Find the critical points by solving \( \frac{dK}{dx} = 0 \). You will find that at \( x^4 = rac{3}{2} \), or \( x = \sqrt[4]{\frac{3}{2}} \). Test the intervals to confirm that this is a maximum.

Key Concepts

First Derivative

To delve into the topic of the first derivative, let's start with the basics. The first derivative of a function gives us incredibly valuable information about its behavior. Specifically, it tells us how the function's output, or its dependent variable, changes in response to changes in its input, the independent variable. In mathematical terms, it's the rate of change or slope of the tangent line to the graph of a function. For our specific example, we are looking at the function given by:

• \( y = \frac{1}{x} \)

Here, we apply the power rule for differentiation, which is a shortcut for finding the derivative of functions of the form \( x^n \). It states that if \( y = x^n \), then the derivative \( y' = nx^{n-1} \). Applying this rule to our function, where \( x^{-1} \) is the form we're working with, the first derivative, \( y' \), becomes:

• \( y' = -\frac{1}{x^2} \)

This result tells us that as you increase \( x \), the rate at which \( y = \frac{1}{x} \) decreases becomes more negative, meaning the curve becomes steeper and steeper downwards as shown in a graph. Understanding this behavior is crucial in identifying how fast changes are happening at any point on the curve.

Second Derivative

The second derivative provides even more depth to our understanding of a function's graph. While the first derivative gives the slope, the second derivative, noted as \( y'' \), informs us about the curvature or concavity. It tells us how the slope is changing, and whether it's increasing or decreasing.

• A positive second derivative means the function is concave up, resembling a smile.
• A negative second derivative indicates the function is concave down, like a frown.

For our function, starting from the first derivative:

• \( y' = -\frac{1}{x^2} \)

We differentiate again to find the second derivative:

• \( y'' = \frac{2}{x^3} \)

Notice that \( y'' \) is always positive for all \( x eq 0 \), confirming that our curve \( y = \frac{1}{x} \) is concave up throughout its domain. This concave-up behavior means the slope of the tangent line is decreasing in absolute value as \( x \) increases.

Curvature Formula

The curvature of a curve at any given point tells us exactly how sharply the curve is bending at that point. To calculate the curvature, we use the curvature formula:

• \( K = \frac{|y''|}{(1 + (y')^2)^{3/2}} \)

Where \( y' \) is the first derivative, providing the slope of the curve, and \( y'' \) is the second derivative, which provides the concavity.
In our example, we've already found:

• \( y' = -\frac{1}{x^2} \)
• \( y'' = \frac{2}{x^3} \)

Plug these into the curvature formula to get:

• \[ K = \frac{2/x^3}{(1 + 1/x^4)^{3/2}} \]

This expression simplifies the understanding of how curvature manifests for the curve \( y = \frac{1}{x} \). As you explore this expression, you see it involves raising combined terms to a power and understanding their influence. Ultimately, this helps identify points of maximum curvature, crucial in many real-world applications such as designing roads or understanding the behavior of physical bodies.