Question

Find the center of mass for \(\rho=\tan ^{2} x\) on \(x \in\left(-\frac{\pi}{4}, \frac{\pi}{4}\right)\).

Short answer

The center of mass is at \( \frac{4}{\pi} - 1 \).

Step-by-step solution

Understand the Problem

To find the center of mass, we must compute the average value of the density function. The function given is \( \rho = \tan^2 x \). We need to find the average value of this function over the interval \( x \in \left(-\frac{\pi}{4}, \frac{\pi}{4}\right) \).

Set Up the Integral for Average Density

The average value of a function \( f(x) \) over an interval \( [a, b] \) is given by \( \frac{1}{b-a} \int_a^b f(x) \, dx \). For our problem, \( a = -\frac{\pi}{4} \), \( b = \frac{\pi}{4} \), and \( f(x) = \tan^2 x \).

Compute the Integral

Evaluate the integral \( \int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \tan^2 x \, dx \). Recall that \( \tan^2 x = \sec^2 x - 1 \), so our integral becomes \( \int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} (\sec^2 x - 1) \, dx \). This can be split into two separate integrals: \( \int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \sec^2 x \, dx - \int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} 1 \, dx \).

Solve the Integrals

First, calculate \( \int \sec^2 x \, dx = \tan x + C \). Thus, \( \int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \sec^2 x \, dx = \tan \left(\frac{\pi}{4}\right) - \tan \left(-\frac{\pi}{4}\right) = 1 - (-1) = 2 \).Now solve \( \int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} 1 \, dx = \left[x\right]_{-\frac{\pi}{4}}^{\frac{\pi}{4}} = \frac{\pi}{4} - (-\frac{\pi}{4}) = \frac{\pi}{2} \).

Combine and Simplify

Subtract the results of the two integrals: \( 2 - \frac{\pi}{2} \). This expression represents the total density over the interval. To find the average density, divide by the length of the interval \( \frac{\pi}{2} \): \( \frac{1}{\frac{\pi}{2}} \times (2 - \frac{\pi}{2}) = \frac{4}{\pi} - 1 \).

Conclusion

The center of mass of the given density function \( \tan^2 x \) over the interval \( x \in \left(-\frac{\pi}{4}, \frac{\pi}{4}\right) \) is found by computing the average density, which is \( \frac{4}{\pi} - 1 \).

Key concepts

Average Value of a Function

When calculating the center of mass, the average value of a function plays a key role. To understand it better, imagine trying to find the average height of people in a room. You would add all their heights and divide by the number of people. Similarly, to find the average value of a continuous function, we use integration.
The average value of a function \( f(x) \) over the interval \( [a, b] \) is computed by the expression:

• \( \frac{1}{b-a} \int_a^b f(x) \, dx \)

Here, \( \int_a^b f(x) \, dx \) sums up all of the area under the curve of \( f(x) \) over the interval \([a, b]\), acting much like adding discrete values.
When dividing by the interval length \((b-a)\), it essentially averages those sums. In our specific problem with \( \tan^2 x \), using this formula helps find the average density across the interval \( x \in \left(-\frac{\pi}{4}, \frac{\pi}{4}\right) \). Understanding this formula is crucial, as it simplifies calculations involving continuous data.

Integral Computation

Integral computation is vital in finding the center of mass. Essentially, integration is like a grand sum that helps compute areas, volumes, and averages. In this problem, we're given \( \rho = \tan^2 x \) and need to calculate

• \( \int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \tan^2 x \, dx \)

To solve it, we use a clever substitution knowing that \( \tan^2 x = \sec^2 x - 1 \). This allows us to break down the integral:

• \( \int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} (\sec^2 x - 1) \, dx = \int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \sec^2 x \, dx - \int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} 1 \, dx \)

Calculating \( \int \sec^2 x \, dx = \tan x \), we apply the limits to find \( 2 \). Similarly, calculating the integral of \( 1 \) gives us \( \frac{\pi}{2} \).
Integral computation doesn't just help us find areas under a curve, but also allows us to simplify complex expressions, breaking them into manageable parts and handling them one at a time.

Density Function

The concept of a density function is essential in many fields, including physics and statistics. It's a way to describe how heavy or concentrated something is over a particular space. In this exercise, our density function \( \rho \) is given by \( \tan^2 x \).
The density function provides the distribution of a material or quantity in a given region. Here, it helps determine how the 'weight' or 'mass' is spread over the interval \( x \in \left(-\frac{\pi}{4}, \frac{\pi}{4}\right) \).
By evaluating the integral of this function, we can understand how much total density there is in this interval. Once we've integrated and divided by the interval's length, we find the average density, helping pinpoint the center of mass.
In exploring the center of mass, the density function guides us in seeing not just where things are most concentrated, but also how they relate spatially, offering a fuller picture of balance and equilibrium across the interval.