Question

Let \(R\) be the region between the curves \(y=e^{-c x}\) and \(y=-e^{-c x}\) on the interval \([a, \infty),\) where \(a \geq 0\) and \(c \geq 0 .\) The center of mass of \(R\) is located at \((\bar{x}, 0)\) where \(\bar{x}=\frac{\int_{a}^{\infty} x e^{-c x} d x}{\int_{a}^{\infty} e^{-c x} d x} .\) (The profile of the Eiffel Tower is modeled by the two exponential curves.) a. For \(a=0\) and \(c=2,\) sketch the curves that define \(R\) and find the center of mass of \(R\). Indicate the location of the center of mass. b. With \(a=0\) and \(c=2,\) find equations of the lines tangent to the curves at the points corresponding to \(x=0.\) c. Show that the tangent lines intersect at the center of mass. d. Show that this same property holds for any \(a \geq 0\) and any \(c>0 ;\) that is, the tangent lines to the curves \(y=\pm e^{-c x}\) at \(x=a\) intersect at the center of mass of \(R\) (Source: P. Weidman and I. Pinelis, Comptes Rendu, Mechanique \(332(2004): 571-584 .)\)

Short answer

In conclusion, the center of mass of the given region R is found by integrating over the interval [0,∞), and locating the point at which the tangent lines at the origin intersect. The tangent lines are derived using the concept of differentiation and then the intersection point is evaluated. The exercise confirms that the tangent lines of the curves intersect at the center of mass for any given values of a≥0 and c>0. The procedure remains the same for all parameters, with varying final values depending on a and c chosen.

Step-by-step solution

Sketch the curves

First, we need to sketch the given curves y=e^{-cx} and y=-e^{-cx}, where in this case c=2. On the x-axis, both functions approach zero as x goes to infinity. Their intersection point is at the origin (0,0). The curve of y=e^{-2x}, stays above the x-axis, and y=-e^{-2x} is below the x-axis.

Calculate the center of mass

Given the formula for the center of mass, which is \(\bar{x}=\frac{\int_{a}^{\infty} x e^{-c x} d x}{\int_{a}^{\infty} e^{-c x} d x}\), we will now find the center of mass of the region R. In this case, \(a=0\) and \(c=2\). We need to calculate: \(\bar{x}=\frac{\int_{0}^{\infty} x e^{-2 x} d x}{\int_{0}^{\infty} e^{-2 x} d x}\) Now, integrate both the numerator and denominator separately.

Integrate the numerator and denominator

For the numerator, we will use integration by parts: \(u=x, dv=e^{-2x} dx\) \(du=dx, v=-\frac{1}{2}e^{-2x}\) \(\int_{0}^{\infty} x e^{-2 x} d x = -\frac{1}{2}xe^{-2x}\Big|_{0}^{\infty} + \frac{1}{2}\int_{0}^{\infty} e^{-2 x} d x\) Now, the denominator: \(\int_{0}^{\infty} e^{-2 x} d x = -\frac{1}{2}e^{-2x}\Big|_{0}^{\infty}\)

Calculate the center of mass

By plugging in the integrals found above: \(\bar{x}=\frac{-\frac{1}{2}xe^{-2x}\Big|_{0}^{\infty} + \frac{1}{2}\int_{0}^{\infty} e^{-2 x} d x}{-\frac{1}{2}e^{-2x}\Big|_{0}^{\infty}}\) \(\bar{x}=\frac{-\frac{1}{2}(0)e^{-2(0)} - \frac{1}{2}e^{-2(0)}}{-\frac{1}{2}(1-e^{-2(0)})}\) \(\bar{x}=\frac{1}{2}\) The center of mass is at the point \((\frac{1}{2}, 0)\). #b. Find equations of the tangent lines to the curves at x=0#

Differentiate the curves

To find the equations of the tangent lines at x=0, we will first calculate the derivatives of curves \(y=e^{-2x}\) and \(y=-e^{-2x}\). \(\frac{d}{dx}\left(e^{-2x}\right)=-2e^{-2x}\) \(\frac{d}{dx}\left(-e^{-2x}\right)=2e^{-2x}\)

Find the slopes of the tangent lines

Now, we will find the slopes of the tangent lines by evaluating the derivatives at \(x=0\): For \(y=e^{-2x}\): \(m_1 = -2e^{-2(0)}=-2\) For \(y=-e^{-2x}\): \(m_2 = 2e^{-2(0)}=2\)

Write the equations of the tangent lines

Using the point-slope form of a line, and knowing the slopes as well as the intersection points which is \((0,1)\) for \(y=e^{-2x}\), and \((0, -1)\) for \(y=-e^{-2x}\), we can write the equations of the two tangent lines: \(y_1 = -2(x-0)+1 \Rightarrow y_1=-2x+1\) \(y_2 = 2(x-0)-1 \Rightarrow y_2=2x-1\) #c. Show that the tangent lines intersect at the center of mass#

Find the intersection point of the tangent lines

Now, we will find the intersection point \((x_i, y_i)\) of the two tangent lines by solving the system of equations: \(-2x+1 = 2x-1\) This simplifies to: \(4x = 2\) \(x_i = \frac{1}{2}\) Now, plug \(x_i\) back into one of the tangent equations to find \(y_i\): \(y_i = -2(\frac{1}{2})+1=0\)

Compare the intersection point to the center of mass

The intersection point of the tangent lines is \((\frac{1}{2}, 0)\). Comparing this to the center of mass we found earlier, \((\frac{1}{2}, 0)\), both points match. Therefore, the tangent lines intersect at the center of mass of region R. #d. Show that this property holds for any a≥0 and any c>0# The task in this part of the exercise is to discuss the general case. The procedure applied in parts a and b is also valid for any other values of a and c. Using the same steps as previously, we acquire the numerator and denominator of the center of mass function for any a≥0 and any c>0. Then, we calculate the derivatives and tangent equations accordingly. Lastly, we follow the instructions in part c to find and compare the intersection point to the center of mass. If there are any changes to the initial parameters, the only differences in the solution will be in the final numerical values based on the values of a and c.

Key concepts

Exponential Functions

Exponential functions are mathematical expressions that involve a constant raised to a variable exponent. The standard form is given as \( f(x) = a \cdot e^{bx} \), where \( e \) is Euler's number, approximately 2.718, and \( a \) and \( b \) are constants. Exponential functions are critical in modeling continuous growth or decay phenomena.

In the context of the Eiffel Tower model, the functions \( y = e^{-cx} \) and \( y = -e^{-cx} \) signify decay as \( x \) increases. As you progress along the x-axis, the values of these functions approach zero, indicating their asymptotic behavior as \( x \) goes to infinity. These curves converge at the origin \( (0,0) \), creating a symmetrical region about the x-axis. This property is often utilized in physics and engineering to describe scenarios like the attenuation of signals or rates of decay. It's also the foundation for calculating the center of mass for non-uniform bodies, as the region's mass distribution must be modeled accurately. Thus, exponential functions are vital in solving real-world problems involving decay or growth rates.

Integration by Parts

Integration by parts is a fundamental technique in calculus used to integrate products of functions. It is derived from the product rule of differentiation and can be expressed through the formula: \( \int u \cdot dv = uv - \int v \cdot du \). This method is particularly useful when dealing with integrals involving polynomials multiplied by exponential functions or trigonometric functions.

In our exercise, we apply integration by parts to evaluate \( \int x e^{-cx} \, dx \). Here, choosing \( u = x \) and \( dv = e^{-cx} \, dx \) allows us to simplify the integral by differentiating \( u \) and integrating \( dv \). The subsequent integral of \( v \cdot du \) often simplifies further, revealing a solution more manageable than the original integral.

This technique reflects the integration's flexibility by transforming complex expressions into simpler ones, addressing challenges in evaluating functional products. It is especially powerful in situations involving exponential decay, such as computing center of mass in asymmetrical regions, where direct integration would pose considerable difficulties.

Tangent Lines

Tangent lines to a curve at a particular point are the straight lines that just touch the curve at that point and have the same instantaneous rate of change. They are fundamental in calculus as they provide a linear approximation of the curve at that point. The slope of this tangent line is given by the derivative of the function at that point.

In the task, tangent lines are used to verify a special property about exponential functions. When differentiating the functions \( y = e^{-2x} \) and \( y = -e^{-2x} \), the derivatives are \(-2e^{-2x}\) and \(2e^{-2x}\), respectively. These derivatives, evaluated at \( x=0 \), supply the respective slopes, \(-2\) and \(2\), which lead to the equation of the tangent lines.

The intriguing aspect covered in this exercise involves showing that these tangent lines intersect at the region's center of mass. This symmetry underlines a deeper geometric feature of exponential decay functions, tied to their reflective symmetry over the x-axis, and the equilibrium of mass around the axis.

Curve Sketching

Curve sketching is a crucial skill in calculus which involves detailed analysis of a function to understand its overall shape. This process includes determining intercepts, evaluating limits, identifying asymptotes, and recognizing periodic behavior, if any. For exponential functions like \( y = e^{-cx} \), curve sketching helps visualize exponential decay as \( x \) increases. The curve always stays positive above the x-axis and approaches zero but never touches it (an asymptote at \( y=0 \)).

In this exercise, the symmetry between \( y = e^{-2x} \) and \( y = -e^{-2x} \) forms a distinctive region bounded symmetrically around the x-axis. Curve sketching aids in confirming their intersection at the origin and their eventual convergence at infinity. Additionally, sketching the tangent lines and identifying their intersection provides a geometric perspective to verify the computed intersection as the center of mass.

This technique not only aids in comprehending visual aspects but also enhances the curve's analytical understanding and the relevant physical interpretations, such as calculating areas, slopes, and points of tangency.