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 \(c>0 .\) 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; see the Guided Project The exponential Eiffel Tower. ) 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 \(R\)

Short answer

Question: Sketch the region R between the curves \(y=e^{-cx}\) and \(y=-e^{-cx}\) for \(c=2\). Find the coordinates of the center of mass \(\bar{x}\) of R for \(a=0\) and \(c=2\) and the equations of the tangent lines at the points corresponding to \(x=0\). Show that the tangent lines intersect at the center of mass, and that this holds true for any \(a \geq 0\) and \(c > 0\). Answer: The center of mass \(\bar{x}\) of R for \(a=0\) and \(c=2\) is located at (1, 0). The tangent lines at \(x=0\) are \(y=-2x+1\) and \(y=2x-1\). The tangent lines intersect at the center of mass, which is \(\left(\frac{1}{2}, 0\right)\) for \(a=0\) and \(c=2\). This intersection holds true for any \(a \geq 0\) and \(c > 0\), as the point of intersection is given by \(\left(a+\frac{1}{c}, 0\right)\), which is the center of mass of R.

Step-by-step solution

a. Sketch the curves and find the center of mass of R

To sketch the curves \(y=e^{-cx}\) and \(y=-e^{-cx}\) for \(c=2\), we observe that both curves are symmetric with respect to the x-axis. For \(x=0\), both curves intersect the y-axis at \((0,0)\). The curve \(y=e^{-2x}\) is decreasing as x increases, and \(y=-e^{-2x}\) is increasing as x increases. Now, to find the center of mass \(\bar{x}\), use the formula given in the exercise: \(\bar{x}=\frac{\int_{a}^{\infty} x e^{-c x} d x}{\int_{a}^{\infty} e^{-c x} d x}\). For \(a=0\) and \(c=2\): \(\bar{x}=\frac{\int_{0}^{\infty} x e^{-2 x} d x}{\int_{0}^{\infty} e^{-2 x} d x}\) = \(\frac{\int_{0}^{\infty} x e^{-2 x} d x}{\int_{0}^{\infty} e^{-2 x} d x}\) To evaluate the integrals, integrate by parts for the numerator: Let \(u=x\) and \(dv=e^{-2x}dx\); then \(du=dx\) and \(v=-\frac{1}{2}e^{-2x}\). \(\int_{0}^{\infty} x e^{-2 x} d x\) = \(\left[-\frac{1}{2}xe^{-2x}|_0^{\infty} \right] + \int_0^\infty \frac{1}{2} e^{-2x}dx\) The limit \(\lim_{x\to\infty} -\frac{1}{2}xe^{-2x}=0\) Therefore, \(\int_{0}^{\infty} x e^{-2 x} d x = \int_0^\infty \frac{1}{2} e^{-2x}dx\) Now evaluate the denominator: \(\int_{0}^{\infty} e^{-2x} dx = -\frac{1}{2}e^{-2x}|_0^{\infty} = \frac{1}{2}\) Finally, compute \(\bar{x}\): \(\bar{x} = \frac{\int_0^\infty \frac{1}{2} e^{-2x} dx}{\frac{1}{2}} = \int_0^\infty \frac{1}{2} e^{-2x} dx = 1\) So, the center of mass of R is located at \((\bar{x}, 0) = (1, 0)\).

b. Find the equations of the tangent lines

To find the equations of the tangent lines at the points corresponding to \(x=0\), we need to find the derivative for each curve \(y=e^{-2x}\) and \(y=-e^{-2x}\). We have: \(y_1 = e^{-2x} \Rightarrow y_1' = -2e^{-2x}\) \(y_2 = -e^{-2x} \Rightarrow y_2' = 2e^{-2x}\) Now, we evaluate the derivative at \(x=0\) for both curves: \(y_1'(0) = -2e^{-2(0)} = -2\) \(y_2'(0) = 2e^{-2(0)} = 2\) The tangent line equations for both curves are: Tangent line at \(y_1\): \(y = -2(x-0) + e^{-2(0)} = -2x + 1\) Tangent line at \(y_2\): \(y = 2(x-0) - e^{-2(0)} = 2x - 1\)

c. Show that the tangent lines intersect at the center of mass

We have the equations for the tangent lines at x=0: Tangent line at \(y_1\): \(y = -2x + 1\) Tangent line at \(y_2\): \(y = 2x - 1\) Now, let's find the point of intersection by setting the tangent lines equal to each other: \(-2x+1 = 2x-1\) Add 2x to both sides: \(1= 4x - 1\) Add 1 to both sides: \(2 = 4x\) Divide by 4: \(x = \frac{1}{2}\) Now we substitute the value of x into one of the tangent line equations to find the corresponding y value: \(y = -2\left(\frac{1}{2}\right) + 1= 0\) So the point of intersection is \(\left(\frac{1}{2}, 0\right)\), which is the center of mass of R for \(a=0\) and \(c=2\).

d. Show that the tangent lines intersect at the center of mass for any \(a \geq 0\) and \(c > 0\)

To show that the tangent lines always intersect at the center of mass, we will first differentiate the general equations \(y = \pm e^{-cx}\): \(y_1 = e^{-cx} \Rightarrow y_1' = -ce^{-cx}\) \(y_2 = -e^{-cx} \Rightarrow y_2' = ce^{-cx}\) Now, we evaluate the derivatives at x=a for both curves: \(y_1'(a) = -ce^{-ca}\) \(y_2'(a) = ce^{-ca}\) The tangent line equations for both curves are: Tangent line at \(y_1\): \(y = -ce^{-c a} (x-a) + e^{-ca}\) Tangent line at \(y_2\): \(y = ce^{-c a} (x-a) - e^{-ca}\) Now, let's find the point of intersection by setting the tangent lines equal to each other: \(-ce^{-c a} (x-a) + e^{-ca} = ce^{-c a} (x-a) - e^{-ca}\) Add \(ce^{-ca} (x-a)\) to both sides: \(e^{-ca} = 2ce^{-c a} (x-a) - e^{-ca}\) Now, we can divide by \(e^{-ca}\) on both sides: \(1 = 2c(x-a)-1\) Add 1 to both sides: \(2 = 2c(x-a)\) Divide by 2c: \(x = a+\frac{1}{c}\) Now substitute x back into one of the tangent lines to find the corresponding y value: \(y = -ce^{-c a} \left(a+\frac{1}{c}-a\right) + e^{-ca}\) \(y = -ce^{-c a} \left(\frac{1}{c}\right) + e^{-ca}\) \(y = -e^{-c a} + e^{-ca} = 0\) So the point of intersection is \(\left(a+\frac{1}{c}, 0\right)\), which is the center of mass of R for any \(a \geq 0\) and \(c > 0\).

Key concepts

Understanding Exponential Functions

Exponential functions are mathematical expressions where a constant base is raised to a variable exponent. In the problem at hand, we encounter the exponential function in the form of y = e^{-cx}, where e is Euler's number (approximately 2.71828), c is a positive constant, and x is the variable.

The function y = e^{-cx} is particularly interesting because it models decay processes in nature and physics, such as radioactive decay and cooling phenomena. This function decreases rapidly toward zero as x increases when c > 0. Visually, its graph is a smooth curve that progressively flattens out and never actually reaches the x-axis, known as asymptotic behavior. The negative exponent indicates the decay effect.

To understand the area between the curves y = e^{-cx} and y = -e^{-cx} from the exercise, we consider the integral of the exponential function over a given interval. This requires knowledge of integral calculus, which we will discuss in the next section. Understanding exponential functions is crucial for grasping the nature of the problem and the behavior of the curves that represent the Eiffel Tower's profile.

Applying Integral Calculus

Integral calculus is the branch of mathematics concerned with the process of integration, which is essentially the reverse operation of differentiation (finding derivatives). It provides us with the tools to calculate areas, volumes, and lengths of curves.

In the context of our problem, integral calculus allows us to find the center of mass of the region R by integrating exponential functions over an interval. The integral of an exponential function like e^{-cx} over the interval [a, ∞), where a >= 0, is a form of improper integral because it extends to infinity.

Technique of Integration by Parts

One effective method to compute such integrals is integration by parts, an integral calculus technique derived from the product rule of differentiation. It is expressed in the formula ∫ u dv = uv - ∫ v du. This technique facilitates the integration of the function x * e^{-cx} by breaking it down into simpler parts. As seen in the step-by-step solution, this process allows the calculation of the center of mass, (̄x, 0), by integrating the moment about the y-axis and then dividing by the total area of R.

Understanding the fundamentals of integral calculus not only supports solving this specific problem but also enables students to approach a broad range of problems in mathematics, physics, and engineering where the calculation of quantities related to areas under curves is required.

Interpreting Tangent Line Equations

Tangent line equations are algebraic representations of lines that just touch a curve at a single point without crossing it. These lines have the same slope as the curve at the point of tangency, which is found by taking the derivative of the curve's equation.

For our exercise, we sought the tangent lines to the curves y = e^{-cx} and y = -e^{-cx} at a particular point, identified by the variable x. By differentiating each curve and evaluating the derivatives at the point of interest, we obtain the slopes of these tangent lines.

We use the slope-point formula to construct the equations of the tangent lines: y - y1 = m(x - x1), where (x1, y1) is the point of tangency and m is the slope. Following this methodology leads us to reveal a fascinating property related to the center of mass: the tangent lines intersect at this point, regardless of the values of a and c.

Understanding this concept is critical in various fields, including physics, where the tangent represents an instantaneous rate of change, and in optimization problems within calculus, where tangent lines signify the slope of a function at maximum or minimum points.