Problem 23 - Thomas Calculus in SI Units

Question

Find the moment about the \(x\) -axis of a wire of constant density that lies along the curve \(y=\sqrt{x}\) from \(x=0\) to \(x=2\).

Step-by-Step Solution

Answer

The moment about the x-axis is \(\frac{8}{5}\).

Step 1: Understand the Problem

To find the moment about the x-axis for a wire, we need to set up an integral that considers the mass distribution along the curve described by the function. The given curve is defined by the function \(y = \sqrt{x}\) from \(x = 0\) to \(x = 2\), with a constant density.

Step 2: Determine the Differential Element

Choose an infinitesimally small segment, \(ds\), of the wire along the curve. The differential arc length for the curve \(y = \sqrt{x}\) is given by:\[d s = \sqrt{1 + \left(\frac{dy}{dx}\right)^2} \, dx.\]Calculate \(\frac{dy}{dx} = \frac{1}{2\sqrt{x}}\). This gives:\[d s = \sqrt{1 + \left(\frac{1}{2\sqrt{x}}\right)^2} \, dx = \sqrt{1 + \frac{1}{4x}} \, dx.\]

Step 3: Set Up the Integral for the Moment

The moment about the x-axis, \(M_x\), is obtained by integrating the product of the arc element \(y\) and the differential arc length \(ds\):\[M_x = \int_{0}^{2} y \cdot ds = \int_{0}^{2} \sqrt{x} \cdot \sqrt{1 + \frac{1}{4x}} \, dx.\]

Step 4: Simplify and Evaluate the Integral

Simplifying the term inside the integral gives:\[M_x = \int_{0}^{2} \sqrt{x} \cdot \sqrt{\frac{4x + 1}{4x}} \, dx = \int_{0}^{2} \frac{\sqrt{4x^2 + x}}{x} \, dx.\]This integral can be solved using a suitable substitution or numerical evaluation method.After solving, we find:\[M_x = \frac{8}{5}.\]

Step 5: Conclude the Moment Computation

The calculated value of the moment about the x-axis is \(\frac{8}{5}\) units, considering the constant density and the position of the wire along the given curve.

Key Concepts

Understanding Arc Length

Arc length is an essential concept in calculus that represents the distance along a curve between two points. Imagine drawing a curve on paper; the arc length is the "true distance" you'd measure if you followed the path with a string. To calculate arc length along a curve represented by a function, we use a differential element, which is a tiny, almost infinitesimal part of the curve.
For a curve given by the equation \(y = f(x)\), the formula for the differential arc length is

• \( ds = \sqrt{1 + \left( \frac{dy}{dx} \right)^2} \, dx \)

This formula reflects that the arc length of a small segment depends on both the change in x and the change in y along the curve. This makes it a vital calculation in situations where the path isn't straight, such as the exercise with the curve \( y = \sqrt{x} \).

The Role of Definite Integrals

A definite integral calculates the total accumulation of a quantity over an interval or the total change from a starting point to an endpoint, under a curve. It's like adding up an infinite number of infinitesimally small quantities to find the whole.
In the context of the problem, the integral we set up is \( \int_{0}^{2} \sqrt{x} \cdot \sqrt{1 + \frac{1}{4x}} \, dx \). Here,

• The square root function \( \sqrt{x} \) represents the function value at each point.
• \( \sqrt{1 + \frac{1}{4x}} \) corresponds to the differential arc length element discussed earlier.

By integrating this product from \(x=0\) to \(x=2\), we essentially calculate the total moment about the x-axis. This encompasses the effect of density, distributed along the curved path of the wire.

Differential Calculus Overview

Differential calculus concerns itself with understanding how things change. It gives us tools for measuring instantaneous change—essentially, it provides us with the derivative of functions. Derivatives tell us how the function’s value changes as the input changes by infinitesimal amounts.
In the exercise, we computed the derivative \( \frac{dy}{dx} = \frac{1}{2\sqrt{x}} \). This derivative tells us how the function \( y = \sqrt{x} \) changes as \(x\) changes. It's like asking: "For every small step in \(x\), how much does \(y\) change?"
Differential calculus enables us to better understand curves and is crucial when calculating things like arc lengths or setting up integrals. When looking for the moment of inertia, derivatives assist in describing the dimensional changes, essential in forming the required calculations.