Problem 134 - Calculus Volume 3

Question

Find the curvature \(\kappa\) for the curve \(y=\frac{1}{3} x^{3}\) at the point \(x=1\).

Step-by-Step Solution

Answer

The curvature \( \kappa \) at \( x=1 \) is \( \frac{\sqrt{2}}{2} \).

Step 1: Recall the formula for curvature

The formula for the curvature \( \kappa \) of a function \( y = f(x) \) is given by:\[ \kappa = \frac{|f''(x)|}{(1 + (f'(x))^2)^{3/2}} \]

Step 2: Find the first derivative

Calculate the first derivative of the function \( y = \frac{1}{3}x^3 \). Using the power rule, we find:\[ f'(x) = x^2 \]

Step 3: Find the second derivative

Now, calculate the second derivative of the function. Differentiating \( f'(x) = x^2 \) again using the power rule gives:\[ f''(x) = 2x \]

Step 4: Substitute the derivatives into the curvature formula

Insert the expressions for \( f'(x) \) and \( f''(x) \) into the curvature formula:\[ \kappa = \frac{|2x|}{(1 + (x^2)^2)^{3/2}} \]

Step 5: Evaluate at the point \(x=1\)

Substitute \( x = 1 \) into the curvature expression:\[ \kappa = \frac{|2\times1|}{(1 + (1^2)^2)^{3/2}} = \frac{2}{(1 + 1)^{3/2}} = \frac{2}{2^{3/2}} = \frac{2}{2\sqrt{2}} = \frac{1}{\sqrt{2}} \]

Step 6: Simplify the result

Simplify \( \frac{1}{\sqrt{2}} \). This can be written as:\[ \frac{1}{\sqrt{2}} = \frac{\sqrt{2}}{2} \] to remove the radical from the denominator.

Key Concepts

Power Rule

The power rule is a straightforward technique in calculus used to find the derivative of a function that is a power of a variable. It particularly comes in handy when working with polynomial functions. To apply the power rule, we look at a term like \(x^n\), where \(n\) is a constant exponent. The formula is:

• If \(y = x^n\), then the derivative \(y' = nx^{n-1}\).

In the original exercise, we applied the power rule to \(y = \frac{1}{3}x^3\). By identifying the term \(x^3\), we use the power rule to find its derivative, yielding \(3\cdot\frac{1}{3}x^{3-1} = x^2\). This simple move efficiently converts a polynomial term into its derivative form, setting the stage for curvature calculation.

Derivative Calculation

Calculating derivatives is essential in determining curvature. Derivatives measure how a function changes as its input changes, like the slope of a tangent to a curve at a given point. There are two derivatives we need for curvature:

• First Derivative: For the function \(y = \frac{1}{3}x^3\), we calculated the first derivative to be \(f'(x) = x^2\). This represents the slope of the tangent line at any point \(x\).
• Second Derivative: The second derivative, \(f''(x) = 2x\), tells us how the slope itself changes, which is crucial for understanding the curvature of the graph.

These derivatives are critical inputs for the curvature formula, providing insights into how the curve bends at any specified point.

Curvature Formula

The curvature (\(\kappa\)) of a function provides a measure of how sharply a curve bends at a particular point. The formula is:

• \(\kappa = \frac{|f''(x)|}{(1 + (f'(x))^2)^{3/2}}\)

When you look at a curve, the curvature at a point often gives insight into how quickly you would change direction if you were moving along it. A higher curvature indicates a sharper turn. In our exercise, the function \(y = \frac{1}{3}x^3\) required substituting the first and second derivatives into this formula. The curvature was calculated at \(x = 1\), which involved plugging \(f'(x) = x^2\) and \(f''(x) = 2x\) into the formula, leading to a curvature of \(\frac{1}{\sqrt{2}}\).

Mathematical Simplification

Simplification is the process of making an expression easier to interpret or use, often by reducing fractions or eliminating radicals. In our original problem, we derived the initial curvature as \(\frac{1}{\sqrt{2}}\). For clarity and ease of use, expressions without square roots in the denominator are often preferred. This is accomplished by multiplying both the numerator and denominator by the square root to "rationalize" the denominator, leading to \(\frac{\sqrt{2}}{2}\).

• This simplification cleans up the expression, often making it neater for further calculations.

These simplifications ensure that results are presented in a standard form, which can be especially helpful for further mathematical analysis or comparison.