Problem 271 - Calculus Volume 1

Question

For the following exercises, compute the center of mass \((\overline{x}, \overline{y}) .\) Use symmetry to help locate the center of mass whenever possible. $$ \begin{array}{l}{ \rho=3 \text { in the triangle with vertices }(0,0), \quad(a, 0)} \\ {\text { and }(0, b)}\end{array} $$

Step-by-Step Solution

Answer

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

Step 1: Understand the Problem

We need to find the center of mass of a triangular region with constant density \( \rho = 3 \) and vertices at \( (0,0), (a,0), (0,b) \). The region is a right triangle.

Step 2: Apply Symmetry Consideration

Since the triangle is a right triangle, its center of mass will be along the median from the vertex at \( (a, 0) \) to \( (0, b) \). The center of mass will lie along the line connecting the midpoint of the hypotenuse and \((a/3, b/3)\), because it's a uniform density triangle.

Step 3: Calculate Center of Mass Formulas

The centroid \((\overline{x}, \overline{y})\) of a triangle with vertices \((x_1, y_1), (x_2, y_2), (x_3, y_3)\) is given by: \[ \overline{x} = \frac{x_1 + x_2 + x_3}{3}, \quad \overline{y} = \frac{y_1 + y_2 + y_3}{3} \] Apply this to the vertices \((0,0), (a,0), (0,b)\).

Step 4: Substitute Vertex Coordinates

Substitute the coordinates of the vertices into the formula: \[ \overline{x} = \frac{0 + a + 0}{3} = \frac{a}{3} \] \[ \overline{y} = \frac{0 + 0 + b}{3} = \frac{b}{3} \]

Step 5: Conclude Center of Mass

So the center of mass \((\overline{x}, \overline{y})\) of the triangle is \( \left( \frac{a}{3}, \frac{b}{3} \right) \). This is consistent with the symmetry considerations.

Key Concepts

Triangle Geometry

Triangle geometry is a fundamental aspect of mathematics that deals with the properties and relations of triangles, which are three-sided polygons. In this specific case, we are looking at a right triangle with vertices at

• (0,0), the origin
• (a,0), on the x-axis
• (0,b), on the y-axis

Each vertex is a point where two sides of the triangle meet. The triangle has a right angle at (0,0), making this geometry a classic example of a right-angled triangle.
One important aspect to consider in triangle geometry is the concept of median lines. A median of a triangle is a line segment joining a vertex to the midpoint of the opposite side. In this right triangle, the centroid or center of mass will lie along the median line, which connects (a,0) and (0,b) with the midpoint of the hypotenuse. Understanding these properties helps us to deduce the line along which the centroid must lie.

Centroid Formula

Finding a triangle's center of mass involves using the centroid formula. The centroid is considered the triangle's balance point, where the triangle could be perfectly balanced on a pinpoint. The formula for finding the centroid

• \( \overline{x} = \frac{x_1 + x_2 + x_3}{3} \)
• \( \overline{y} = \frac{y_1 + y_2 + y_3}{3} \)

demonstrates the average of the vertices' x and y coordinates. This formula gives a very simple way to compute the coordinates of the centroid as long as the vertices' coordinates are known.
When we plug in the vertices

• (0,0),
• (a,0),
• (0,b)

we get the calculated centroid coordinates as \( \left( \frac{a}{3}, \frac{b}{3} \right) \). Applying this formula shows the immense value of a standardized approach in geometric problems involving triangles.

Symmetry in Geometry

Symmetry plays a crucial role in geometry and can help in easily locating important points such as centroids. A shape is said to be symmetric when one half is a mirror image of the other. With respect to the given triangle

• (0,0),
• (a,0),
• (0,b)

there is inherent symmetry in the arrangement of the triangle's sides, particularly considering it is a right triangle.
The centroid of a triangle with uniform density, like this one, is found on what can be imagined as the 'balance line', derived from the symmetry. In this problem, the centroid lies on the line with symmetry through the median. This simplifies the process of determining the center of mass greatly, as these symmetrical properties allow certain predictable geometric calculations, aiding in reducing complexity to mere arithmetic. By understanding symmetry, one can also predict behavior and solve for other geometric properties in more complex polygons or three-dimensional shapes.