Problem 71 - Single Variable Calculus

Question

Find the centroid of the region under the graph of \(f(x)=\cosh x\) on \([-a, a]\)

Step-by-Step Solution

Answer

The centroid of the region under the graph of \(f(x)=\cosh x\) on \([-a, a]\) is at the point \((\frac{1}{2}, a - 1)\).

Step 1: To find the area of the region under the curve, integrate the function \(f(x) = \cosh x\) over the interval \([-a, a]\). The area formula is given by: \[A = \int_{-a}^a f(x) dx\] #Step 2: Compute the area of the region under the curve#

To find the area of the region under the curve, we must compute the integral: \[A = \int_{-a}^a \cosh x\, dx\] Since the derivative of the function \(\sinh x\) is \(\cosh x\), we can integrate: \[A = \left[\sinh x \right]_{-a}^a\] Now, evaluate the definite integral by substituting the bounds: \[A = \sinh a - \sinh (-a)\] Since \(\sinh (-x) = - \sinh x\), we get: \[A = \sinh a + \sinh a = 2 \sinh a\] #Step 3: Compute the moments about the axes#

Step 2: To find the moments about the respective axes, we must integrate the product of the distance from the axis and the function over the interval \([-a, a]\). For the moment about the \(x\)-axis (\(M_x\)), we have: \[M_x = \int_{-a}^a x f(x) dx\] Similarly, for the moment about the \(y\)-axis (\(M_y\)), we have: \[M_y = \frac{1}{2}\int_{-a}^a f(x) dx\] Now, we must compute the integrals: \[M_x = \int_{-a}^a x \cosh x\, dx\] Using integration by parts: Let \(u=x\) and \(dv=\cosh x dx\) Then, \(du=dx\) and \(v= \sinh x\) Using the integration by parts formula, we get: \[M_x = \left[x \sinh x\right]_{-a}^a - \int_{-a}^a \sinh x\, dx\] Now, evaluate the definite integral by substituting the bounds: \[M_x = \left[a \sinh a + (-a) \sinh(-a) \right] - (\sinh a - \sinh (-a))\] Since \(\sinh (-x) = - \sinh x\), we get: \[M_x = 2a \sinh a - 2 \sinh a\] \[M_y = \frac{1}{2} (2 \sinh a) = \sinh a\] #Step 4: Find the coordinates of the centroid#

To find the coordinates of the centroid, use the formulas \(\bar{x} = \frac{M_y}{A}\) and \(\bar{y} = \frac{M_x}{A}\): \[\bar{x} = \frac{M_y}{A} = \frac{\sinh a}{2 \sinh a} = \frac{1}{2}\] \[\bar{y} = \frac{M_x}{A} = \frac{2a \sinh a - 2 \sinh a}{2 \sinh a} = a - 1\] Thus, the centroid of the region under the graph of \(f(x) = \cosh x\) on \([-a, a]\) is at the point \((\frac{1}{2}, a - 1)\).

Key Concepts

Definite Integral

The definite integral is a fundamental concept in calculus, representing the accumulation of a quantity over an interval. This mathematical tool allows us to calculate the area under a curve bounded by the graph of a function within specific limits.

When dealing with a function like \( f(x)=\text{cosh} x \), which describes a hyperbolic cosine curve, the definite integral is used to find the total area under this curve between two points, \(-a\) and \(a\). In the context of finding the centroid, the definite integral not only provides the area but is also part of the process in determining the moments about the x-axis and y-axis, key steps in locating the centroid of a region.

Hyperbolic Functions

Understanding Hyperbolic Sine and Cosine

Hyperbolic functions, such as the hyperbolic sine \(\sinh(x)\) and hyperbolic cosine \(\cosh(x)\), are analogs of the trigonometric functions but for hyperbola rather than a circle. These functions can describe the shape of hanging cables or chains, called catenaries, and have useful properties similar to trigonometric functions.

For instance, the derivative of \(\sinh(x)\) is \(\cosh(x)\) and vice versa. Moreover, they exhibit qualities such as \(\sinh(-x) = -\sinh(x)\), which simplifies calculations involving symmetry, especially when integrating over a symmetric interval like \([-a, a]\). These properties are particularly beneficial when determining areas under curves and calculating centroids of symmetric regions.

Centroid Formulas

The Balance Point of a Region

Centroids are analogous to the center of mass or balance point of a geometric region. They are calculated using moments, which in the context of geometry, are integrals that represent the weighted average position of the area under a curve.

Mathematically stated, the centroid \((\bar{x}, \bar{y})\) of a region can be found using the formulas \(\bar{x} = \frac{M_y}{A}\) and \(\bar{y} = \frac{M_x}{A}\), where \(M_x\) and \(M_y\) are the moments about the x- and y-axes respectively, and \(A\) is the area of the region. These formulas encapsulate the essence of balance and symmetry in a shape, contributing to a wide range of applications from engineering to physics.

Integration By Parts

A Technique for Complex Functions

Integration by parts is a technique derived from the product rule in differentiation and is used to integrate products of functions. The formula is given by \(\int u dv = uv - \int v du\), where \(u\) and \(dv\) are functions of x.

This technique becomes particularly useful when faced with a product of two functions that are not straightforward to integrate together, as is the case when finding moments in centroid calculations. For the exercise provided, integration by parts simplifies the process of integrating \(x\cosh x\), allowing for the determination of the moment about the x-axis, and subsequently, the y-coordinate of the centroid.