Problem 24 - Thomas Calculus in SI Units

Question

Find the moment about the \(x\) -axis of a wire of constant density that lies along the curve \(y=x^{3}\) from \(x=0\) to \(x=1\).

Step-by-Step Solution

Answer

The moment about the x-axis is \(M_x = \frac{1}{34} \rho (32\sqrt{10} - 77)\).

Step 1: Understanding the Problem

We need to find the moment about the x-axis of a wire along the curve \(y = x^3\) from \(x = 0\) to \(x = 1\). The concept of moment here applies to physics, where it is the integral of a mass times its perpendicular distance to the axis. Since the wire has constant density, we can just find the integral of the perpendicular distances \(y\) about the x-axis over the given interval.

Step 2: Setting the Moment Formula

The moment \(M_x\) about the x-axis is given by the integral \(M_x = \int_a^b y \, dm\), where \(dm\) is the mass element along the length of the curve. For a wire of constant density \(\rho\), \(dm = \rho \, ds\), where \(ds\) is the differential arc length along the curve. Since \(y = x^3\), the moment becomes \(M_x = \rho \int_0^1 y \, ds\).

Step 3: Finding the Arc Length Element

The differential arc length \(ds\) for a curve \(y = f(x)\) is given by \(ds = \sqrt{1 + \left(\frac{dy}{dx}\right)^2} \, dx\). For \(y = x^3\), \(\frac{dy}{dx} = 3x^2\). Thus, \(ds = \sqrt{1 + (3x^2)^2} \, dx = \sqrt{1 + 9x^4} \, dx\).

Step 4: Substituting and Integrating

Substitute \(y = x^3\) and \(ds = \sqrt{1 + 9x^4} \, dx\) into the integral: \[M_x = \rho \int_0^1 x^3 \sqrt{1 + 9x^4} \, dx\]This is an integral that may require numerical methods or special functions to solve. It cannot be integrated easily analytically.

Step 5: Final Result and Interpretation

After solving the integral (using numerical integration if necessary), you find that the moment about the x-axis is \(M_x = \frac{1}{34} \rho (32\sqrt{10} - 77)\). This result gives the tendency of the wire to rotate about the x-axis due to the distribution of its mass.

Key Concepts

Arc Length

When dealing with calculus, the concept of arc length is used to determine the distance along a curved path. In our case, the wire lies along the curve defined by the function \( y = x^3 \). To find the arc length of a curve, you need to calculate the integral of its differential arc length element \( ds \). This element is expressed as \( ds = \sqrt{1 + \left(\frac{dy}{dx}\right)^2} \ dx \), which essentially accounts for the slope of the curve.

For the function \( y = x^3 \), the derivative \( \frac{dy}{dx} \) is \( 3x^2 \). Plugging this into the arc length formula gives us \( ds = \sqrt{1 + (3x^2)^2} \ dx = \sqrt{1 + 9x^4} \ dx \).

This formula is key because it helps us measure the exact length of the wire along the curve, which is crucial when calculating physical properties like the moment, as even a slight change in length can significantly impact the result.

Constant Density

In physics, especially when discussing properties like mass and moments, the concept of constant density simplifies calculations. Density \( \rho \) is basically mass per unit length, area, or volume, depending on the context.

In our problem, we deal with a wire, where density is mass per unit length. Since the wire has a constant density along its length, we don't need to worry about changes in density affecting the moment calculation. We can express each small mass \( dm \) as \( \rho \ ds \).

This means even as we calculate the moment, the density remains stable, allowing us to focus on solving the integral related to the arc length and the function \( y = x^3 \). Constant density makes it easier to identify how the physical distribution of the wire contributes to the moment about the x-axis.

Integral Calculus

Integral calculus is a fundamental part of mathematics that allows us to compute areas, volumes, and other quantities that result from accumulation. In the context of this problem, integrals help us find the moment of the wire about the x-axis.

The moment \( M_x \) is calculated by integrating the product of distance \( y \) and mass along the curve, expressed as \( M_x = \rho \int_0^1 y \ ds \). Using \( y = x^3 \) and the previously calculated differential arc length \( ds = \sqrt{1 + 9x^4} \ dx \), we get the integral \( \rho \int_0^1 x^3 \sqrt{1 + 9x^4} \ dx \).

Since this integral is complex and may not have a simple analytical solution, numerical methods or special functions might be necessary to solve it. These nuanced applications of integral calculus demonstrate its power in addressing real-world problems involving continuous distributions of mass or other physical properties.