Question

A thin plate is bounded by the graphs of \(y=e^{-x}, y=-e^{-x}, x=0,\) and \(x=L .\) Find its center of mass. How does the center of mass change as \(L \rightarrow \infty ?\)

Short answer

As the length of the thin plate approaches infinity, the coordinates of the center of mass converge to (1, 1/2).

Step-by-step solution

Find the area of the region

To find the area of the region, we can use the formula \(A = \int_{a}^{b} [f(x) - g(x)] dx\). Plugging in our functions and limits of integration, we get: $$ A = \int_{0}^{L} [e^{-x} - (-e^{-x})] dx $$ Now we can integrate and find the area. $$ A = 2 \int_{0}^{L} e^{-x} dx $$ $$ A = 2 [-e^{-x}] \Big|_0^L = 2(1 - e^{-L}) $$

Find the center of mass coordinates

Now that we have the area, we can find the coordinates of the center of mass, starting with \(\bar{x}\): $$ \bar{x} = \frac{1}{A}\int_{0}^{L} x[e^{-x} - (-e^{-x})] dx $$ $$ \bar{x} = \frac{1}{2(1 - e^{-L})} \int_{0}^{L} 2xe^{-x} dx $$ Here we can use integration by parts with \(u = x\) and \(dv = 2e^{-x} dx\). Then \(du = dx\) and \(v = -2e^{-x}\), so we have: $$ \bar{x} = \frac{1}{2(1 - e^{-L})}[-2xe^{-x}\Big|_0^L - (-2) \int_0^L e^{-x} dx] $$ By evaluating the integral, we get: $$ \bar{x} = \frac{1}{2(1 - e^{-L})}[-2Le^{-L}\big|_0^L - (-2)(1 - e^{-L})] $$ $$ \bar{x} = \frac{1}{2(1 - e^{-L})}[2L e^{-L} - 2 + 2e^{-L}] $$ Now, we can find \(\bar{y}\): $$ \bar{y} = \frac{1}{A}\int_{0}^{L} \frac{1}{2}(e^{-2x} - (-e^{-x})^2) dx $$ $$ \bar{y} = \frac{1}{2(1 - e^{-L})}\int_{0}^{L} (e^{-2x}) dx $$ Now, integrate and find \(\bar{y}\). $$ \bar{y} = \frac{1}{2(1 - e^{-L})}[-\frac{1}{2}e^{-2x}\Big|_0^L = \frac{1}{2(1 - e^{-L})} \left(1 - \frac{ e^{-2L}}{2}\right) $$

Analyze the center of mass as \(L \rightarrow \infty\)

Now, we will analyze how the center of mass coordinates change as \(L \rightarrow \infty\). Starting with \(\bar{x}\): $$ \lim_{L \rightarrow \infty} \bar{x} = \lim_{L \rightarrow \infty} \frac{1}{2(1 - e^{-L})}[2L e^{-L} - 2 + 2e^{-L}] $$ Using L'Hôpital's rule, we get \(\lim_{L \rightarrow \infty} \frac{2L e^{-L}}{2 - 2e^{-L}} = 1\). Therefore, as \(L \rightarrow \infty\), the x-coordinate of the center of mass converges to 1. Now, let's analyze \(\bar{y}\): $$ \lim_{L \rightarrow \infty} \bar{y} = \lim_{L \rightarrow \infty} \frac{1}{2(1 - e^{-L})} \left(1 - \frac{ e^{-2L}}{2}\right) $$ Notice that as \(L \rightarrow \infty\), \(e^{-2L} \rightarrow 0\), so the y-coordinate of the center of mass converges to \(\frac{1}{2}\). The center of mass of the thin plate changes as \(L \rightarrow \infty\) to the coordinates \((1, \frac{1}{2})\).

Key concepts

Integration

Integration is a fundamental tool in calculus that helps us find areas under curves. In the context of finding the center of mass, integration allows us to calculate the total area and mass distribution of an object. By looking at the area between two curves, we can use definite integrals to determine exact values.
For instance, consider the function difference between two curves, such as \( f(x) - g(x) \). To find the area under this section, we use:\[ A = \int_{a}^{b} [f(x) - g(x)] \ dx \]In the case of the thin plate problem, we are dealing with the exponential functions \( e^{-x} \) and \( -e^{-x} \).

• First, set up the integral over the desired interval. In our example, it's from 0 to \( L \).
• Add the two curves because they are symmetric about the x-axis.
• Use the integral to find the area, which involves calculating: \( 2 \int_{0}^{L} e^{-x} \ dx \).
• This evaluates to: \( 2(1 - e^{-L}) \), which gives the bounded area that contributes to mass calculation.

Integration by Parts

Integration by parts is a technique that helps solve integrals that are products of functions, especially when straightforward integration isn't feasible. It is based on the product rule for differentiation and is given by:\[ \int u \ dv = uv - \int v \ du \]When solving for \( \bar{x} \), we encounter the integral \( \int_0^L 2xe^{-x} \ dx \). To tackle this, consider:

• Let \( u = x \), which means \( du = dx \).
• Choose \( dv = 2e^{-x} \ dx \), resulting in \( v = -2e^{-x} \).

Putting it together, we apply integration by parts:
\[ \bar{x} = \frac{1}{2(1 - e^{-L})}[-2xe^{-x}\Big|_0^L - (-2) \int_0^L e^{-x} \ dx] \]
This technique simplifies the integration process, leading us to the x-coordinate of the center of mass. It's crucial to express products of functions in a form that allows integration by parts to be effective.

Limit as L Approaches Infinity

Analyzing the behavior of functions as a variable approaches a limit is called evaluating a limit. Here, we examine how the center of mass coordinates behave as \( L \to \infty \). This principle is significant when considering infinite or large regions in calculus.

• For \( \bar{x} \), the tricky part involves applying L'Hôpital's Rule as both numerator and denominator approach zero: \( \lim_{L \to \infty} \frac{2L e^{-L}}{2 - 2e^{-L}} = 1 \).
• For \( \bar{y} \), the exponential term \( e^{-2L} \) approaches zero, simplifying the evaluation.
• At infinity, the mass distribution balances such that the centroid tends towards specific coordinates.

The result is that the center of mass of the system converges to a stable position: at \((1, \frac{1}{2})\). Understanding the concept of limits is critical in determining how functions and coordinates stabilize or diverge in extended domains.