Problem 58 - Calculus III

Question

The surface of a large cup is formed by revolving the graph of the function \(y=0.25 x^{1.6}\) from \(x=0\) to \(x=5\) about the \(y\) -axis (measured in centimeters).Find the curvature \(\kappa\) of the generating curve as a function of \(x\).

Step-by-Step Solution

Answer

The curvature \(\kappa(x)\) is \(\frac{0.24}{x^{0.4} (1 + 0.16 x^{1.2})^{3/2}}\).

Step 1: Find the Derivative of y

Given the function \(y = 0.25 x^{1.6}\), we need to find its first derivative \(\frac{dy}{dx}\). Using the power rule for derivatives, \(\frac{dy}{dx} = 0.25 \cdot 1.6 \cdot x^{1.6 - 1} = 0.4 x^{0.6}\).

Step 2: Find the Second Derivative of y

Next, find the second derivative \(\frac{d^2y}{dx^2}\) by differentiating \(\frac{dy}{dx} = 0.4 x^{0.6}\) again. Using the power rule, \(\frac{d^2y}{dx^2} = 0.4 \cdot 0.6 \cdot x^{0.6 - 1} = 0.24 x^{-0.4}\).

Step 3: Curvature Formula

The formula for the curvature \(\kappa\) of a curve given by \(y = f(x)\) is \( \kappa = \frac{|y''|}{(1 + y'^2)^{3/2}} \). We will use this formula to calculate \(\kappa\) for our given function.

Step 4: Substitute the Derivatives into the Curvature Formula

Substitute \(\frac{d^2y}{dx^2} = 0.24 x^{-0.4}\) and \(\frac{dy}{dx} = 0.4 x^{0.6}\) into the curvature formula: \[ \kappa = \frac{|0.24 x^{-0.4}|}{(1 + (0.4 x^{0.6})^2)^{3/2}} = \frac{0.24 x^{-0.4}}{(1 + 0.16 x^{1.2})^{3/2}} \].

Step 5: Simplify the Expression for Curvature

Simplify the expression to find \(\kappa\): Thus, \(\kappa(x) = \frac{0.24}{x^{0.4} (1 + 0.16 x^{1.2})^{3/2}}\).

Key Concepts

Revolving Surfaces

A revolving surface is an intriguing concept in geometry and calculus, where a two-dimensional curve is spun around an axis to generate a three-dimensional object. In this particular exercise, we are dealing with a surface created by revolving the function \( y = 0.25 x^{1.6} \) around the \( y \)-axis.When we revolve a curve, each point traces out a circle around the axis, and all these circles together form a surface. This is a crucial process used in various applications like designing cups, bowls, and other symmetrical shapes.

• The original curve provides the shape of the object when rotated.
• The axis of rotation determines the dimension and symmetry of the surface.
• For functions revolving around the \( y \)-axis, each circle’s radius is based on the \( x \)-coordinate of the point.

Understanding revolving surfaces is important for visualizing how flat, two-dimensional representations produce complex three-dimensional forms when rotated, a foundational concept in calculus and engineering.

Calculus Derivatives

In calculus, derivatives play a fundamental role in understanding the behavior of functions. A derivative represents how a function changes as its input changes, which is crucial for analyzing rates of change and the shape of curves. In this exercise, we calculated the first and second derivatives of the function \( y = 0.25 x^{1.6} \).

First Derivative

The first derivative, \( \frac{dy}{dx} \), measures the slope and rate of change of the function at any point \( x \). For our function, applying the power rule, we find \( \frac{dy}{dx} = 0.4 x^{0.6} \). This tells us how steep the curve is at different points.

• The power rule simplifies finding derivatives for functions of the form \( ax^n \).
• Understanding the slope of a curve helps anticipate how the function behaves as \( x \) increases.

Second Derivative

The second derivative, \( \frac{d^2y}{dx^2} \), provides information on how the slope itself changes, which helps in understanding the concavity or curvature of the function.

• For our example, \( \frac{d^2y}{dx^2} = 0.24 x^{-0.4} \).
• A positive second derivative indicates the curve is concave up, while a negative indicates concave down.

This knowledge is key to exploring the curve's geometry and further analyzing functions for applications such as finding maximum and minimum values.

Curvature of Curves

Curvature quantifies how sharply a curve bends at any given point. It is especially important in understanding the behavior of curves in space. In this exercise, we computed the curvature of the curve defined by the function \( y = 0.25 x^{1.6} \).The formula for curvature \( \kappa \) is:\[ \kappa = \frac{|y''|}{(1 + (y')^2)^{3/2}} \]Using this, the curvature helps describe the curve’s "twistiness," which is important for applications in physics and engineering where path or trajectory needs precision.

Calculating Curvature

To determine the curvature, we substitute the first derivative \( \frac{dy}{dx} = 0.4 x^{0.6} \) and the second derivative \( \frac{d^2y}{dx^2} = 0.24 x^{-0.4} \) into the curvature formula:\[ \kappa(x) = \frac{0.24}{x^{0.4} (1 + 0.16 x^{1.2})^{3/2}} \]

• This result shows us how the shape of the curve changes in different parts of its domain.
• A larger curvature value indicates a tighter curve, while a smaller value indicates a gentler bend.

Understanding the curvature of curves is essential for fields that require precise modeling of paths, such as road and railway design, and for more complex tasks in mechanical engineering.