Problem 15 - Calculus Early Transcendentals: Pearson New International Edition

Question

Find the center of mass of the homogeneous triangle with vertices \((a, 0,0),(0, a, 0)\), and \((0,0, a)\).

Step-by-Step Solution

Answer

The center of mass of the triangle is \(\left(\frac{a}{3}, \frac{a}{3}, \frac{a}{3}\right)\).

Step 1: Identify the Coordinates of the Vertices

Begin by identifying the coordinates of the vertices of the triangle. The given vertices of the triangle are \((a, 0, 0)\), \((0, a, 0)\), and \((0, 0, a)\).

Step 2: Understand the Formula for the Center of Mass

The center of mass (centroid) for a triangle is found by taking the arithmetic mean (average) of the coordinates of its vertices. The formula is:\[(x_{cm}, y_{cm}, z_{cm}) = \left(\frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3}, \frac{z_1 + z_2 + z_3}{3}\right)\]where \((x_1, y_1, z_1)\), \((x_2, y_2, z_2)\), and \((x_3, y_3, z_3)\) are the coordinates of the vertices.

Step 3: Apply the Coordinates into the Formula

Substitute the coordinates of the vertices into the formula for the centroid:- For \(x\)-coordinates: \(x_{cm} = \frac{a + 0 + 0}{3} = \frac{a}{3}\)- For \(y\)-coordinates: \(y_{cm} = \frac{0 + a + 0}{3} = \frac{a}{3}\)- For \(z\)-coordinates: \(z_{cm} = \frac{0 + 0 + a}{3} = \frac{a}{3}\)

Key Concepts

Centroid of a Triangle

In geometry, the centroid of a triangle is a point that is considered the 'center' of the triangle. It is sometimes referred to as the center of mass because it balances the triangle equally in all directions, assuming the triangle has a uniform density. To find the centroid of a triangle, we use a straightforward formula that involves the average of the vertices' coordinates.

The centroid is important because it tells us about the distribution of the triangle's mass. While methods for finding centroids vary depending on the shape, for a triangle, it always involves averaging the coordinates.

This consists of finding the arithmetic mean of the x-coordinates, the y-coordinates, and the z-coordinates of the three vertices. The result gives you a single coordinate, the centroid, which acts as the triangle's center of gravity.

Why does this work? In short, it's because averaging the coordinates essentially balances out all varying distances from the edge to the center.

Coordinates of Vertices

The coordinates of vertices of a triangle are key to identifying many of its properties, including the centroid. Vertices are simply the corners or intersection points of the triangle's sides.

In a 3-dimensional space, each vertex is represented by a set of coordinates, usually in \(x, y, z\) format. Each value corresponds to the position along the designated Cartesian axis.

For our particular triangle, the vertices are located at \(a, 0, 0\), \(0, a, 0\), and \(0, 0, a\). This means each vertex lies distinctly on one of the axes, each distance being \(a\) from the origin in its respective axis.

• The vertex \(a, 0, 0\) is \(a\) units along the x-axis.
• The vertex \(0, a, 0\) is \(a\) units along the y-axis.
• The vertex \(0, 0, a\) is \(a\) units along the z-axis.

The coordinates are essential inputs for calculating the centroid using the simple averaging technique.

Arithmetic Mean

The arithmetic mean is a central concept used to calculate the centroid of a triangle. It involves taking the sum of numbers and dividing by the count of numbers. In the context of finding a triangle's centroid, it provides an average coordinate value.

Math symbols often represent this process. For example, if you are finding the x-coordinate of the centroid, you add all the x-values from the vertices and divide by 3. The same process applies for both the y and z coordinates. Thus, the centroid formula reads:

• \(x_{cm} = \frac{x_1 + x_2 + x_3}{3}\)
• \(y_{cm} = \frac{y_1 + y_2 + y_3}{3}\)
• \(z_{cm} = \frac{z_1 + z_2 + z_3}{3}\)

This simple average calculation effectively provides a balanced central point that considers the triangle's full volume and symmetry in 3-dimensional space. It explains why each coordinate of the centroid lies at \(\frac{a}{3}\) for our given triangle. Using the arithmetic mean ensures that the centroid is truly at the center of mass rather than an arbitrary point in space.