Problem 12 - Thomas Calculus in SI Units

Question

Find the center of mass of a thin plate of constant density \(\delta\) covering the given region. The region bounded above by the curve \(y=1 / x^{3}\), below by the curve \(y=-1 / x^{3},\) and on the left and right by the lines \(x=1\) and \(\bar{x}=a>1 .\) Also, find \(\lim _{a \rightarrow \infty} \bar{x}\).

Step-by-Step Solution

Answer

The center of mass is \(\bar{x} = \frac{1 - \frac{1}{a}}{1 - \frac{1}{a^2}}\), and \(\lim_{a \to \infty} \bar{x} = 1\).

Step 1: Identify the Region

The given region is bounded by the curves \(y = \frac{1}{x^3}\) and \(y = -\frac{1}{x^3}\) between \(x = 1\) and \(x = a\) where \(a > 1\). This region is symmetric with respect to the x-axis.

Step 2: Calculate the Area of the Region

The area of the region can be calculated by integrating the difference of the upper curve and lower curve:\[ A = \int_{1}^{a} \left( \frac{1}{x^3} - \left(-\frac{1}{x^3}\right) \right) \, dx = \int_{1}^{a} \frac{2}{x^3} \, dx = 2 \left[ -\frac{1}{2x^2} \right]_{1}^{a} = \left( 1 - \frac{1}{a^2} \right). \]

Step 3: Find the Center of Mass (ar{x})

The horizontal coordinate of the center of mass, \(\bar{x}\), is given by:\[ \bar{x} = \frac{1}{A} \int_{1}^{a} x \left(\frac{1}{x^3} - \left(-\frac{1}{x^3}\right)\right) \, dx. \]Substitute the area and simplify:\[ \bar{x} = \frac{1}{1 - \frac{1}{a^2}} \int_{1}^{a} \frac{2}{x^2} \, dx = \frac{1}{1 - \frac{1}{a^2}} \cdot 2 \left[ -\frac{1}{x} \right]_{1}^{a} = \frac{1}{1 - \frac{1}{a^2}} \left( 1 - \frac{1}{a} \right). \]

Step 4: Find the Limit \(\lim_{a \to \infty} \bar{x}\)

Evaluate the limit of \(\bar{x}\) as \(a \to \infty\):\[ \lim_{a \to \infty} \bar{x} = \lim_{a \to \infty} \frac{1}{1 - \frac{1}{a^2}} \left( 1 - \frac{1}{a} \right) = \lim_{a \to \infty} \frac{1 - \frac{1}{a}}{1 - \frac{1}{a^2}}. \]As \(a\) goes to infinity, \(-\frac{1}{a}\) and \(-\frac{1}{a^2}\) both approach zero, thus:\[ \lim_{a \to \infty} \bar{x} = 1. \]

Key Concepts

Calculus

Calculus is an essential branch of mathematics that deals with the study of change and motion. It provides tools for analyzing dynamic systems and calculating quantities that vary continuously. Two fundamental concepts of calculus are differentiation and integration, each serving unique purposes in mathematical analysis.
While differentiation focuses on the rate of change, integration concerns itself with the accumulation of quantities over an interval. Calculus enables us to solve complex problems involving curves and motion in various scientific fields.

• Differentiation involves finding the slope of a curve at a point, which is useful in understanding how functions change.
• Integration, on the other hand, deals with finding the total accumulation between two points, such as areas under curves.
• Mastering the basics of calculus is key to solving advanced mathematical problems, especially those involving limits and center of mass calculations.

Understanding these concepts lays the groundwork for tackling more complex challenges, such as finding the center of mass for a given region.

Integration

Integration is a powerful tool in calculus used to ascertain the cumulative value of functions over a specific interval, often represented graphically by the area under curves. This process is crucial in physics and engineering for calculating total quantities, such as displacement or area.
When integrating, we effectively reverse the process of differentiation, piecing together small, infinitesimal parts to understand a whole. This is especially useful when dealing with functions that describe real-world phenomena.

• Definite integrals provide the exact area between a function and the x-axis over an interval \(\[a, b\]\).
• The integral of a function is often noted as \[\int f(x) \, dx\].
• To compute complex integrals, techniques such as substitution and integration by parts might be applied.

For the exercise at hand, integration helps determine the area of the region between curves and further calculate the center of mass by finding weighted average positions over that area.

Limits

Limits are an essential concept in calculus, helping us understand the behavior of functions as they approach specific points or infinity. By evaluating limits, mathematicians can deduce the values a function approaches, but never actually reaches, which can be vital in understanding the continuity and the end behavior of functions.
Limits are especially powerful when dealing with indefinite values or when a function becomes undefined at certain points.

• To calculate a limit as \(x\rightarrow a\) means finding the value that the function approaches as \(x\) gets closer to \(a\).
• Limits are closely connected to the derivatives, as they define the slope of tangent lines to functions.
• In the context of the exercise, limits help determine how the center of mass \(\bar{x}\) behaves as the parameter \(a\) extends to infinity.

Through the application of limits, we can find out that as \(a\) approaches infinity, the calculated center of mass reaches a stable value.

Center of Mass Calculation

The center of mass, or \(\bar{x}\), is a pivotal concept in physics and engineering, representing the average position of a distributed mass. For a plate of constant density, calculating the center of mass requires integrating across the region to find weighted average positions.
You balance these positions to understand where the mass concentrates, providing insights into the equilibrium and stability of the object.

• The horizontal coordinate of center of mass \(\bar{x}\) is given by formula \[\bar{x} = \frac{1}{A} \int x \, (density) \, dx\].
• This formula is applied with the area \(A\) calculated from integrating across the defined region.
• In symmetrical regions, like in the exercise, the symmetry can simplify calculations by balancing the contributions from equal parts.

The problem’s solution shows how to integrate and use limits efficiently to find \(\bar{x}\), illustrating that math can solve practical problems in assessing balance and equilibrium.