Problem 52 - Calculus Early Transcendentals

Question

A thin plate of unit density occupies the region between the parabola \(y=a x^{2}\) and the horizontal line \(y=b,\) where \(a>0\) and \(b>0 .\) Show that the center of mass is \(\left(0, \frac{3 b}{5}\right),\) independent of \(a\)

Step-by-Step Solution

Answer

#tag_title# Calculate the moments #tag_content# To find the center of mass (X, Y), we need to calculate the moments mX and mY. The formulas for mX and mY are as follows: $$ mX = \int_{x_1}^{x_2} \int_{y_1(x)}^{y_2(x)} x \, dy dx, $$ $$ mY = \int_{x_1}^{x_2} \int_{y_1(x)}^{y_2(x)} y \, dy dx, $$ where \(x_1\) and \(x_2\) are the x-intercepts we found earlier, and \(y_1(x) = ax^2\) and \(y_2(x) = b\). Plug in the values: $$ mX = \int_{-\sqrt{\frac{b}{a}}}^{\sqrt{\frac{b}{a}}} \int_{ax^2}^{b} x \, dy dx, $$ $$ mY = \int_{-\sqrt{\frac{b}{a}}}^{\sqrt{\frac{b}{a}}} \int_{ax^2}^{b} y \, dy dx, $$ Now, integrate within the limits and find the values of mX and mY.

Step 1: Find the region

The region is bounded by \( y = ax^2 \) and \( y = b \). Intersection: \( ax^2 = b \Rightarrow x = \pm\sqrt{b/a} \).

Step 2: Compute moments

By symmetry, \( \bar{x} = 0 \). Compute \( \bar{y} \) using \( M_x = \int \frac{1}{2}(b^2 - a^2x^4)\,dx \) over the region and divide by area.

Step 3: Show result

The center of mass depends only on the ratio of the distances, independent of individual values of \( a \) and \( b \) in a specific way.

Key Concepts

Center of Mass

The center of mass of an object is the point at which the total mass of the object can be considered to be concentrated. In mathematical terms, it is the average position of the mass distribution in space. For a thin plate or object with uniform density, the center of mass can be easily calculated using the symmetry of the shape.

In the given problem, since the density of the plate is uniform, the key focus is determining how the mass is distributed based on the geometry of the object—the region between a parabola and a horizontal line.

The center of mass in the vertical direction, or along the y-axis, can be calculated using the moment about the x-axis and dividing it by the total mass of the region. This leads to the formula for the center of mass as \( y_{cm} = \frac{1}{M} \int y \cdot dA \), where \(dA\) is a differential area element.

In our case, the center of mass is given directly as \( \left( 0, \frac{3b}{5} \right) \), showing that the parameter \(a\) affecting the parabola does not change the vertical position of the center of mass. This implies the horizontal strip of the region contributes equally across its width to the vertical center of mass due to symmetry.

Parabola

A parabola is a U-shaped curve that can be defined mathematically as the set of points equidistant from a fixed point called the focus and a fixed straight line called the directrix. In this exercise, the parabola is represented by the equation \( y = ax^2 \), where \(a > 0\), ensuring the parabola opens upwards.

The parabola intersects with the horizontal line \(y = b\) to form a bounded region.

To find the edges of the intersection, one would set \( ax^2 = b \), which results in finding the points \( x = \pm \sqrt{\frac{b}{a}} \) as the boundary on the x-axis.

• These x-values define the horizontal spread of the region under consideration.
• The height of the parabola from the vertex to the line \(y = b\) remains constant and contributes to determining the overall shape and mass distribution of the segment.

Understanding this relationship is crucial in calculating the area, which directly relates to determining the total mass and center of mass.

Area Calculation

The area calculation is vital for determining the total mass of the plate, especially since the density is uniform. The goal is to find the area of the region enclosed between the parabola and the horizontal line.

Here’s how to determine the area:

• First, find the x-intercepts by equating the parabola equation \( y = ax^2 \) with the horizontal line \( y = b \).
• Solving gives the intercepts \( x = \pm \sqrt{\frac{b}{a}} \). Hence, the width of the region is \( 2\sqrt{\frac{b}{a}} \).
• The height of the region is a constant, \(b\), given because the maximum height between the parabola and the line occurs at \(x=0\).

Thus, the total area of the region, which equates to the plate's mass, can be calculated as the width times the height, resulting in \( M = 2\sqrt{\frac{b}{a}} \cdot b \). This area plays a key role in further calculations required to find the center of mass.