Problem 273 - Calculus Volume 1

Question

For the following exercises, use a calculator to draw the region, then compute the center of mass \((\overline{x}, \overline{y}) .\) Use symmetry to help locate the center of mass whenever possible. The region bounded by \(\quad y=\cos (2 x)\) \(x=-\frac{\pi}{4}, \quad\) and \(x=\frac{\pi}{4}\)

Step-by-Step Solution

Answer

The center of mass is \((\overline{x}, \overline{y}) = (0, \frac{\pi}{8})\).

Step 1: Sketch the Region

First, we sketch the curve \(y = \cos(2x)\) on a graph for the interval \(-\frac{\pi}{4} \leq x \leq \frac{\pi}{4}\). The cosine function has a period of \(\pi\), but here it's compressed by a factor of 2, so it completes one full wave over \([-\frac{\pi}{2}, \frac{\pi}{2}]\). Evaluate key points: at \(x = -\frac{\pi}{4}\), \(x = 0\), and \(x = \frac{\pi}{4}\) to draw the curve. The graph is symmetrical about the y-axis.

Step 2: Recognize Symmetry

Observe that the function \(y = \cos(2x)\) is symmetric about the y-axis. Thus, the region bounded by the curve and the vertical lines \(x = -\frac{\pi}{4}\) and \(x = \frac{\pi}{4}\) will have its center of mass on the y-axis, which implies \(\overline{x} = 0\).

Step 3: Compute the Area of the Region

The total area \(A\) under the curve from \(x = -\frac{\pi}{4}\) to \(x = \frac{\pi}{4}\) is computed using the integral: \[ A = \int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \cos(2x) \, dx. \] Evaluate this integral to find \(A\). The antiderivative of \(\cos(2x)\) is \(\frac{1}{2} \sin(2x)\), so: \[ A = \left[ \frac{1}{2} \sin(2x) \right]_{-\frac{\pi}{4}}^{\frac{\pi}{4}} = \frac{1}{2}(\sin(\frac{\pi}{2}) - \sin(-\frac{\pi}{2})) = 1. \]

Step 4: Calculate \(\overline{y}\) (Center of Mass)

Using symmetry for the x-coordinate, \(\overline{x} = 0\). To find \(\overline{y}\), use the formula: \[ \overline{y} = \frac{1}{A} \int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} y(x) \cdot \cos(2x) \, dx. \] Since \(y(x)\) is \(\cos(2x)\) itself in this context: \[ \overline{y} = \frac{1}{1} \int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \cos^2(2x) \, dx. \] Use the identity \(\cos^2(2x) = \frac{1 + \cos(4x)}{2}\). Substitute and integrate: \[ \overline{y} = \frac{1}{2} \left[ x + \frac{1}{4} \sin(4x) \right]_{-\frac{\pi}{4}}^{\frac{\pi}{4}} = \frac{1}{2} \left( \frac{\pi}{4} - (-\frac{\pi}{4}) \right) = \frac{\pi}{8}. \]

Step 5: State the Center of Mass

The center of mass of the region is \((\overline{x}, \overline{y}) = (0, \frac{\pi}{8})\). This takes into account the symmetry center at the y-axis and the averaging over the y-values determined by the integral.

Key Concepts

Symmetry in Calculus

Symmetry is a crucial concept in calculus that can greatly simplify calculations. It arises when a function exhibits a mirrored behavior, making certain calculations more straightforward. In our exercise, the function \( y = \cos(2x) \) showcases symmetry about the y-axis. This characteristic means that for every positive \( x \), there exists an equivalent negative \( x \), such that the function values are equal. This mirror image effect makes it easier to find the center of mass.When we talk about symmetry in calculus, consider the following key points:

• Symmetry about the y-axis implies that \( f(x) = f(-x) \).
• This symmetry allows us to deduce that the center of mass along the x-axis, denoted \( \overline{x} \), is zero, as the positive and negative halves balance each other out.
• Understanding and recognizing symmetry reduces the complexity of integration and other calculus operations.

Emphasizing symmetry can lead to significant insights when calculating areas, volumes, or describing physical phenomena, such as balancing forces.

Integrals in Calculus

Integrals are fundamental in calculus, serving as tools for area and central values calculation. The integral of a function gives the total accumulated value, often corresponding to physical quantities like the area under a curve.In this exercise, the integral \[A = \int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \cos(2x) \, dx\]represents the area under the curve of \( y = \cos(2x) \). Understanding how to evaluate this integral is key to finding the center of mass:

• The process involves finding the antiderivative, which for \( \cos(2x) \) results in \( \frac{1}{2} \sin(2x) \).
• By evaluating this antiderivative at the given limits, we obtained the area \( A \). Calculating this accurately was crucial for resolving the center of mass \( \overline{y} \).
• The calculated integral helps us establish the weight that different parts of the curve add to the whole, influencing the determination of the center of mass.

Overall, mastering integrals helps unravel complex shapes and forms, offering clarity to otherwise intricate calculus problems.

Trigonometric Functions

Trigonometric functions like sine and cosine are periodic functions that repeat their values in regular intervals and are extremely useful in calculus. Our given function, \( y = \cos(2x) \), belongs to this family and follows a clear pattern which is essential when dealing with periodic regions.Understanding trigonometric functions involves noting the following:

• The cosine function, \( \cos(2x) \), compressed by a factor of 2, completes a full wave over the interval \(-\frac{\pi}{2} \text{ to } \frac{\pi}{2}\).
• Trigonometric identities aid in simplifying the integrals. Consider the identity \( \cos^2(2x) = \frac{1 + \cos(4x)}{2} \), which was used to find \( \overline{y} \) in the exercise.
• By exploiting these properties, we can more easily navigate through complex curves and deduce essential qualities like symmetry and periodicity.

Recognizing and using these trigonometric principles enables you to smoothly approach a range of calculus tasks and apply mathematical theory to real-life rotations and oscillations.