Problem 154 - Calculus 3

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). [T] Use technology to graph the curvature function.

Step-by-Step Solution

Answer

Graph the curvature function of the surface using the second derivative of the function.

Step 1: Understand the Problem

We need to find the parametric representation of the surface when revolving the given function around the y-axis. The function provided is \(y = 0.25x^{1.6}\), and the range for \(x\) is from 0 to 5.

Step 2: Set Up the Revolution

To find the surface of revolution, we will revolve the graph around the y-axis. This means for each point \((x, y)\) on the graph, we need to consider its rotation in 3D space about the y-axis. The element of surface area is often calculated using the formula for the surface of revolution.

Step 3: Calculate the Surface Area Element

The surface area \(A\) of a revolution about the y-axis can be calculated by the formula: \[ A = 2\pi \int_{a}^{b} x \sqrt{1 + \left(\frac{dy}{dx}\right)^2} \, dx \] First, we need to find \(\frac{dy}{dx}\) for \(y = 0.25x^{1.6}\).

Step 4: Differentiate the Function

Find the derivative of the function \(y = 0.25x^{1.6}\). Using the power rule: \[ \frac{dy}{dx} = 0.25 \times 1.6 x^{0.6} = 0.4 x^{0.6} \]

Step 5: Substitute the Derivative

Substitute \(\frac{dy}{dx}\) into the surface area formula: \[ A = 2\pi \int_{0}^{5} x \sqrt{1 + (0.4 x^{0.6})^2} \, dx \] This integral can be evaluated using technology or by numerical methods.

Step 6: Graph the Curvature Function

Now let's graph the curvature function. The curvature \(k\) of a curve \(y = f(x)\) can be calculated as \[ k = \frac{|f''(x)|}{(1 + (f'(x))^2)^{3/2}} \] Find the second derivative \(f''(x)\) and substitute it into the curvature formula. Use graphing software to plot the curvature over the interval \(x = 0 \text{ to } 5\).

Key Concepts

Curvature Function

The curvature function of a curve guides how much the curve is bending at any given point. It's a measure of how sharply a curve changes direction, which is pivotal in understanding the overall geometry of a surface that might result from such a curve. In mathematical terms, the curvature \( k \) of a function \( y = f(x) \) is given by the formula:

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

This equation uses both the first and second derivatives of the function. The first derivative \( f'(x) \) indicates the slope or the rate of change, while the second derivative \( f''(x) \) measures the rate of change of the rate of change, which translates into curvature. By plotting this function, you get a visual insight into how the curve bends over a specified interval. For the function \( y = 0.25x^{1.6} \), computing and graphing the curvature reveals how it behaves from \( x = 0 \) to \( x = 5 \), providing critical data for further analyses.

Parametric Representation

When dealing with surfaces of revolution, parametric representation becomes crucial. It offers a convenient way to describe curves and surfaces in three-dimensional space. The original curve is represented by parameters rather than traditional \( x \) and \( y \) coordinates. For the exercise, the function \( y = 0.25x^{1.6} \) revolves around the \( y \)-axis. To parametrize this, consider:

• \( x = t \)
• \( y = 0.25t^{1.6} \)
• \( z = 0 \)

These parametric equations transform the original curve into a 3D spiral, helping visualize its revolution. This transformation is crucial for calculating geometric properties like surface area. Parametric representation, thus, aids in both visualizing and quantifying surfaces of revolution, making it a powerful tool in calculus.

Differentiation

Differentiation is a core technique in calculus, used to find the derivative of a function. The derivative represents the rate of change and is fundamental in many mathematical applications, including the calculation of surfaces of revolution. In our exercise, we start with the function \( y = 0.25x^{1.6} \). To differentiate, apply the power rule:

• \( \frac{dy}{dx} = 0.25 \times 1.6 \times x^{0.6} = 0.4 x^{0.6} \)

This derivative is pivotal for both evaluating the integral used in calculating the surface area and for determining the curvature of the function. Differentiation opens the door to deeper insights into the behavior of functions, highlighting instantaneous rates of change and enabling complex analyses like finding areas or optimizing processes.

Numerical Integration

Numerical integration is a powerful technique used when an integral cannot be solved analytically, which is often the case with complex or non-standard functions. In this exercise, the goal is to find the surface area generated by revolving the curve around the \( y \)-axis. The surface area \( A \) is calculated by:

• \[ A = 2\pi \int_{0}^{5} x \sqrt{1 + (0.4 x^{0.6})^2} \, dx \]

Solving this integral analytically could be challenging, so numerical methods provide an effective solution. Methods like the trapezoidal rule, Simpson’s rule, or software tools help approximate the value, allowing for a practical calculation of the surface area. Numerical integration thus complements analytical techniques, ensuring that even when traditional calculus fails, approximations give meaningful results.