Problem 14 - Calculus of a Single Variable

Question

Find \(M_{x}, M_{y},\) and \((\overline{x}, \overline{y})\) for the laminas of uniform density \(\boldsymbol{\rho}\) bounded by the graphs of the equations. \(y=6-x, y=0, x=0\)

Step-by-Step Solution

Answer

The moments of the lamina are \(M_x = 18 \rho\) and \(M_y = 18 \rho\) and the centroid of the lamina is located at (2, 2).

Step 1: Identify the Dimensions of the Triangle

The vertices of the right triangle formed by the equations \(y = 6 - x\), \(y = 0\), and \(x = 0\) are at (0, 0), (6, 0), and (0, 6). Hence, this is a right triangle with a base and height of 6 units.

Step 2: Compute for the Area A of the Triangle

Use the formula for area of a triangle \(A = 1/2 * base * height\). So, \(A = 1/2 * 6 * 6 = 18\).

Step 3: Compute for the Moments Mx and My

Given a uniform density, \(\rho\), \(M_x\) and \(M_y\) are computed as \(M_y = \rho * 1/2 * base^2\) and \(M_x = \rho * 1/2 * height^2\). Hence, \(M_x = \rho * 1/2 * 6^2 = 18 \rho\) and \(M_y = \rho * 1/2 * 6^2 = 18 \rho\).

Step 4: Find the Centroid (\(\overline{x}\), \(\overline{y}\))

The centroid of a triangle is located at \((1/3) * base\) from \(x = 0\) and \((1/3) * height\) from \(y = 0\). Therefore, \((\overline{x}, \overline{y}) = (6/3, 6/3) = (2, 2)\).

Key Concepts

Centroid

The centroid of a triangle is its center of mass, where the triangle balances perfectly if made of a uniform material. For the right triangle in our exercise, the equations given provide a shape with vertices at (0, 0), (6, 0), and (0, 6). To find the centroid, we use basic principles of geometry.

• The centroid can be calculated using the formula: \( \left( \overline{x}, \overline{y} \right) = \left( \frac{1}{3} \text{base}, \frac{1}{3} \text{height} \right). \)
• This gives us the position \(\left(2, 2\right)\) for our triangle, indicating that the centroid is located 2 units from both the x and y-axes, perfectly situating it within the triangle's bounds.

Understanding the position of the centroid is crucial as it helps in various physical applications, such as determining moments of inertia.

Lamina

A lamina refers to a thin flat layer or sheet, especially in a structural context. In this exercise, the lamina is the physical representation of our triangle, a two-dimensional shape.

• Since the lamina has uniform density, its physical properties, like mass and distribution, are constant across its surface.
• The moments \(M_x\) and \(M_y\), which measure how the mass is distributed with respect to the axes, are key concepts when analyzing a lamina. Moment calculations consider both the geometric and material properties of the lamina.

In engineering and physics, this understanding of laminae is fundamental for predicting how objects will react under various forces.

Uniform Density

Uniform density implies that every point in the lamina has the same density. This means each unit of area has the same amount of mass.

• In calculations, density (symbolized by \(\rho\)) ensures that the mass and geometry are proportionally reflected in moments and centroid computations.
• Given that the triangle we are analyzing has a uniform density, the calculations for moments \(M_x\) and \(M_y\) become straightforward, since density \(\rho\) acts as a constant multiplier. Both moments value result in \(18 \rho\) due to the consistent density throughout the shape.

Uniform density simplifies the mathematical modeling of physical problems, allowing us to generalize about the behavior of such shapes.

Right Triangle

In geometry, a right triangle is a triangle where one angle is exactly 90 degrees. The right triangle from our exercise is bounded by the lines \(y = 6 - x\), \(y = 0\), and \(x = 0\), forming its hypotenuse, base, and height, respectively.

• This triangle is described fully by its two sides (base and height) both being equal to 6 units, given the vertex points (0, 0), (6, 0), and (0, 6).
• The relationship of the sides allows us to calculate the area and subsequently, essential characteristics like the centroid and moments of inertia.

Recognizing a right triangle's properties within coordinate systems aids in simplifying many complex geometrical problems, providing a solid base for analytical calculations.