Problem 144 - Calculus 3

Question

Find the radius of curvature of \(6 y=x^{3}\) at the point \(\left(2, \frac{4}{3}\right)\)

Step-by-Step Solution

Answer

The radius of curvature at the point is \( \frac{5\sqrt{5}}{2} \).

Step 1: Find the First Derivative

Start by expressing the equation implicitly as \( 6y = x^3 \). Differentiate both sides with respect to \( x \). We get \( 6\frac{dy}{dx} = 3x^2 \), so \( \frac{dy}{dx} = \frac{x^2}{2} \).

Step 2: Find the Second Derivative

Differentiate \( \frac{dy}{dx} = \frac{x^2}{2} \) with respect to \( x \) again to obtain the second derivative. This gives us \( \frac{d^2y}{dx^2} = x \).

Step 3: Radius of Curvature Formula

The formula for the radius of curvature \( R \) at a particular point is given by:\[R = \frac{\left( 1 + \left( \frac{dy}{dx} \right)^2 \right)^{3/2}}{\left| \frac{d^2y}{dx^2} \right|}\]

Step 4: Evaluate the Derivatives at the Point

Evaluate \( \frac{dy}{dx} \) at the point \( x = 2 \). Substituting \( x = 2 \) into \( \frac{dy}{dx} = \frac{x^2}{2} \), we get \( \frac{dy}{dx} = 2 \). Similarly, substitute \( x = 2 \) into \( \frac{d^2y}{dx^2} = x \), obtaining \( \frac{d^2y}{dx^2} = 2 \).

Step 5: Calculate Radius of Curvature

Substitute the values obtained into the formula for \( R \):\[R = \frac{\left( 1 + (2)^2 \right)^{3/2}}{|2|} = \frac{\left( 1 + 4 \right)^{3/2}}{2} = \frac{\sqrt{5}^3}{2}\]Simplifying further, \( R = \frac{5\sqrt{5}}{2} \).

Key Concepts

First Derivative

The first derivative of a function, often represented as \( \frac{dy}{dx} \), is essentially how we measure the rate of change or the slope of a function at a particular point. When we have an equation like \( 6y = x^3 \), finding the first derivative helps us understand how \( y \) changes with respect to small changes in \( x \). In the given problem, we start by differentiating both sides of the equation \( 6y = x^3 \).

• The derivative of \( 6y \) with respect to \( x \) gives \( 6 \frac{dy}{dx} \), since we are treating \( y \) as a function of \( x \).
• On the right side, the derivative of \( x^3 \) is \( 3x^2 \).

By equating these results \( 6 \frac{dy}{dx} = 3x^2 \), we solve for \( \frac{dy}{dx} \) to get \( \frac{dy}{dx} = \frac{x^2}{2} \). This tells us how the function's slope changes as \( x \) changes at any point, showing the immediate rate of change of \( y \).
This initial step is crucial because it sets the groundwork for finding how the curve is behaving at various points.

Second Derivative

The second derivative, represented as \( \frac{d^2y}{dx^2} \), gives us an idea of the curvature or concavity of the function. Essentially, while the first derivative tells us about the slope, the second derivative tells us how that slope itself changes as we move along the curve. From the problem, after finding that \( \frac{dy}{dx} = \frac{x^2}{2} \), we differentiate this expression again with respect to \( x \) to find the second derivative.

• Differentiating \( \frac{x^2}{2} \) results in \( \frac{d^2y}{dx^2} = x \).

This tells us how the rate of change of the slope behaves. If \( \frac{d^2y}{dx^2} \) is positive, the curve is concave up at that point, and if it is negative, the curve is concave down. Understanding the second derivative is particularly important when determining curvature because it affects how sharply or gently the function curves.

Implicit Differentiation

Implicit differentiation is a technique used when you cannot easily solve a function for one variable, like in equations where \( y \) is not isolated. Instead of solving for \( y \) beforehand, implicit differentiation allows us to differentiate both sides of an equation without explicitly solving for \( y \).In our example, the equation \( 6y = x^3 \) doesn't easily allow for solving \( y \) in terms of \( x \) completely. Implicitly differentiating:

• We differentiate both sides: The left side is differentiated as \( 6 \frac{dy}{dx} \), using the chain rule (since \( y \) is a function of \( x \)).
• The right side, \( x^3 \), differentiates to \( 3x^2 \).

This technique is powerful because it allows us to find the derivatives needed to analyze and understand the structure of a curve, like its radius of curvature, even when dealing with non-standard or complex equations. Implicit differentiation is a key part of advanced calculus and is invaluable for handling a wide array of mathematical problems where direct solutions aren't practical.