Problem 14 - Calculus Early Transcendentals

Question

Mass of one-dimensional objects Find the mass of the following thin bars with the given density function. $$\rho(x)=\left\\{\begin{array}{ll} 1 & \text { if } 0 \leq x \leq 2 \\ 2 & \text { if } 2 < x \leq 3 \end{array}\right.$$

Step-by-Step Solution

Answer

Answer: The total mass of the thin bar is 4 units.

Step 1: Find Mass of Interval 1

To find the mass in Interval 1, we need to integrate the density function over the interval \(0 \leq x \leq 2\). Since the density function is constant (\(\rho(x)=1\)), the integral will simply be the length of the interval: $$m_1 = \int_{0}^{2} \rho(x) dx = \int_{0}^{2} 1 dx$$ Now, we determine the mass by integrating: $$m_1 = [x]_{0}^{2} = (2-0) = 2$$

Step 2: Find Mass of Interval 2

To find the mass in Interval 2, we need to integrate the density function over the interval \(2 < x \leq 3\). Since the density function is constant (\(\rho(x)=2\)), the integral will simply be twice the length of the interval: $$m_2 = \int_{2}^{3} \rho(x) dx = \int_{2}^{3} 2 dx$$ Now, we determine the mass by integrating: $$m_2 = [2x]_{2}^{3} = 2(3-2) = 2$$

Step 3: Calculate Total Mass

We have found the mass in each interval, and to get the total mass of the bar, we need to sum up the masses of both intervals: $$m_{total} = m_1 + m_2 = 2 + 2 = 4$$ The total mass of the thin bar is 4 units.

Key Concepts

Mass Calculus

Mass calculus is an essential concept in physics and mathematics concerned with determining the mass of an object through integration techniques. When dealing with one-dimensional objects, such as thin bars, the mass is calculated by integrating the density function of the object over its length.

This allows us to sum up the continuous distribution of density over the entire object, providing a precise measurement of total mass. For the given problem of a thin bar with different densities in specific intervals, we calculate the mass separately for each unique density region and then sum them up.

The concept is especially useful for engineering and physical problems, making it possible to predict how structures will react under various conditions.

Density Function

A density function, in the context of mass calculus, describes how mass is distributed along a one-dimensional object. The density function \(\rho(x)\) gives us the mass per unit length at any point along the object.

In the exercise provided, the density function changes at a specific point. It is vital to divide the bar into segments based on where the density function \(\rho(x)\) varies. For example:

• For \(0 \leq x \leq 2\), the density \(\rho(x)\) remains constant at 1.
• For \(2 < x \leq 3\), the density \(\rho(x)\) remains constant at 2.

The process involves understanding the piecewise nature of the function to correctly integrate and find the mass of each segment. This understanding helps in accurately capturing the mass variations across different parts of the object.

Integration Techniques

Integration techniques are the mathematical methods used to calculate the mass of an object from its density function. In our exercise, we use basic integration to find the mass of each interval of the thin bar.

By integrating the density function over specified intervals, we sum up the contributions to mass from tiny segments, providing the total mass of that part of the bar. The steps in this process include:

• Setting up the integral based on the density function's definition in each region.
• Calculating the definite integral over the specified bounds for each section of the bar.
• Finally, adding the results from different intervals to get the total mass.

These techniques make it straightforward to handle problems involving continuous variable densities and highlight the power of calculus in practical applications.